Mixing
Solve : If a, b, c, d are in H.P. , then prove that ab + bc + cd = 3ad
$begingroup$
My try : Since a, b, c, d are in HP, 1/a, 1/b, 1/c and 1/d are in AP.
If the common difference of AP is k, then 1/d = 1/a + 3k
==> k = (1/3)(1/d - 1/a) = (1/3){(a - d)/ad} = (a-d)/3ad
and 1/c = 1/a + 2k = 1/a + 2(a-d)/3ad = [3d + 2a - 2d]/3ad = (2a + d)/3ad
==> c = 3ad/(2a + d)
Further, a, b, c are in HP, ==> b = 2ac/(a+c)
So, ab + bc = b(a+c) = 2ac/(a+c) = 2ac
ab + bc + cd = 2ac + cd = c(2a+d)
Substituting for c from step (ii) above,
ab + bc + cd = {3ad/(2a+d)}*(2a+d) = 3ad [Proved]
I think that this process is quite long , is there any another approach of this question.
sequences-and-series
asked Sep 9 '18 at 4:01
user580093user580093
11316
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add a comment |
$begingroup$
My try : Since a, b, c, d are in HP, 1/a, 1/b, 1/c and 1/d are in AP.
If the common difference of AP is k, then 1/d = 1/a + 3k
==> k = (1/3)(1/d - 1/a) = (1/3){(a - d)/ad} = (a-d)/3ad
and 1/c = 1/a + 2k = 1/a + 2(a-d)/3ad = [3d + 2a - 2d]/3ad = (2a + d)/3ad
==> c = 3ad/(2a + d)
Further, a, b, c are in HP, ==> b = 2ac/(a+c)
So, ab + bc = b(a+c) = 2ac/(a+c) = 2ac
ab + bc + cd = 2ac + cd = c(2a+d)
Substituting for c from step (ii) above,
ab + bc + cd = {3ad/(2a+d)}*(2a+d) = 3ad [Proved]
I think that this process is quite long , is there any another approach of this question.
sequences-and-series
asked Sep 9 '18 at 4:01
user580093user580093
11316
$endgroup$
$begingroup$
Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Sep 9 '18 at 7:24
add a comment |
$begingroup$
My try : Since a, b, c, d are in HP, 1/a, 1/b, 1/c and 1/d are in AP.
If the common difference of AP is k, then 1/d = 1/a + 3k
==> k = (1/3)(1/d - 1/a) = (1/3){(a - d)/ad} = (a-d)/3ad
and 1/c = 1/a + 2k = 1/a + 2(a-d)/3ad = [3d + 2a - 2d]/3ad = (2a + d)/3ad
==> c = 3ad/(2a + d)
Further, a, b, c are in HP, ==> b = 2ac/(a+c)
So, ab + bc = b(a+c) = 2ac/(a+c) = 2ac
ab + bc + cd = 2ac + cd = c(2a+d)
Substituting for c from step (ii) above,
ab + bc + cd = {3ad/(2a+d)}*(2a+d) = 3ad [Proved]
I think that this process is quite long , is there any another approach of this question.
sequences-and-series
asked Sep 9 '18 at 4:01
user580093user580093
11316
$endgroup$
My try : Since a, b, c, d are in HP, 1/a, 1/b, 1/c and 1/d are in AP.
If the common difference of AP is k, then 1/d = 1/a + 3k
==> k = (1/3)(1/d - 1/a) = (1/3){(a - d)/ad} = (a-d)/3ad
and 1/c = 1/a + 2k = 1/a + 2(a-d)/3ad = [3d + 2a - 2d]/3ad = (2a + d)/3ad
==> c = 3ad/(2a + d)
Further, a, b, c are in HP, ==> b = 2ac/(a+c)
So, ab + bc = b(a+c) = 2ac/(a+c) = 2ac
ab + bc + cd = 2ac + cd = c(2a+d)
Substituting for c from step (ii) above,
ab + bc + cd = {3ad/(2a+d)}*(2a+d) = 3ad [Proved]
I think that this process is quite long , is there any another approach of this question.
sequences-and-series
sequences-and-series
asked Sep 9 '18 at 4:01
user580093user580093
11316
asked Sep 9 '18 at 4:01
user580093user580093
11316
asked Sep 9 '18 at 4:01
user580093user580093
11316
asked Sep 9 '18 at 4:01
user580093user580093
11316
asked Sep 9 '18 at 4:01
user580093user580093
11316
11316
$begingroup$
Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Sep 9 '18 at 7:24
add a comment |
$begingroup$
Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Sep 9 '18 at 7:24
$begingroup$
Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Sep 9 '18 at 7:24
$begingroup$
Please read this MathJax tutorial, which explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Sep 9 '18 at 7:24
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since $a$, $b$, $c$, $d$ are in HP, $c^{-1}+a^{-1}=2b^{-1}$ and $d^{-1}-b^{-1}=c^{-1}-a^{-1}$.
begin{align}
(c^{-1}+a^{-1})(d^{-1}-b^{-1})&=2b^{-1}(c^{-1}-a^{-1}) \
(cd)^{-1}+(ad)^{-1}-(ab)^{-1}-(bc)^{-1}&=2(bc)^{-1}-2(ab)^{-1} \
(cd)^{-1}+(ad)^{-1}+(ab)^{-1}&=3(bc)^{-1} \
(abcd)[(cd)^{-1}+(ad)^{-1}+(ab)^{-1}]&=(abcd)[3(bc)^{-1}] \
ab+bc+cd&=3ad \
end{align}
answered Sep 9 '18 at 4:30
YutaYuta
62929
$endgroup$
$begingroup$
Your solution is very similar to what OP has done. I think OP wants an alternate method which is shorter.
$endgroup$
– Anurag A
Sep 9 '18 at 4:39
add a comment |
$begingroup$
To organize your method:
$$k=frac13left(frac1d-frac1aright)=frac12left(frac1c-frac1aright)=frac1b-frac1a,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$.
Equating $(1,3),(2,3),(1,2)$ we get:
$$ab=3ad-2bd\
bc=2ac-ab\
cd=3ad-2ac\
ab+bc+cd=3ad-2bd+2ac-ab+3ad-2ac=\
=3ad+(3ad-2bd)-ab=3ad.$$
Alternative method:
$$k=frac1b-frac1a=frac1c-frac1b=frac1d-frac1c,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$. It results in:
$$ab=frac{a-b}{k}; bc=frac{b-c}{k}; cd=frac{c-d}{k};\
ab+bc+cd=frac{a-b+b-c+c-d}{k}=frac{a-d}{k}=3ad,$$
because:
$$frac{a-d}{k}=3ad iff k=frac13left(frac1d-frac1aright).$$
answered Sep 9 '18 at 6:54
farruhotafarruhota
21.8k2842
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add a comment |
$begingroup$
This is no quicker, but for what it’s worth:
If $a,b,c,d$ are in harmonic progression, then $frac1a,frac1b,frac1c, frac1d$ are in arithmetic progression, so $frac1a+frac1d=frac1b+frac1c$, or (just adding the fractions) $$displaystyle frac{a+d}{color{red}{ad}}=frac{b+c}{bc}.$$
Also, $frac1a+frac1c=frac2b$ and $frac1b+frac1d=frac2c$, or $frac{a+c}{ac}=frac2b$ and $frac{b+d}{bd}=frac2c$, from which $$displaystylefrac{2a+2c}{color{green}{abc}}=frac4{b^2}mbox{ and }displaystylefrac{2b+2d}{color{blue}{bcd}}=frac4{c^2}$$.
From this, $displaystyle color{red}{ad}=frac{(a+d)bc}{b+c}=frac{abc+bcd}{b+c}$, and you therefore need to show that $displaystyle ab+bc+cd=3frac{abc+bcd}{b+c}$, or equivalently that $$displaystyle (b+c)(ab+bc+cd)=3(abc+bcd).$$
Now, $3(abc+bcd)\=(abc+bcd)+2(color{green}{abc}+color{blue}{bcd})\=abc+bcd+(ab^2+b^2c)+(bc^2+c^2d)\=ab^2+b^2c+bcd+abc+bc^2+c^2d\=b(ab+bc+cd)+c(ab+bc+cd).$
edited Jan 29 at 7:23
user69284
17918
answered Sep 9 '18 at 19:23
Steve KassSteve Kass
11.4k11530
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $a$, $b$, $c$, $d$ are in HP, $c^{-1}+a^{-1}=2b^{-1}$ and $d^{-1}-b^{-1}=c^{-1}-a^{-1}$.
begin{align}
(c^{-1}+a^{-1})(d^{-1}-b^{-1})&=2b^{-1}(c^{-1}-a^{-1}) \
(cd)^{-1}+(ad)^{-1}-(ab)^{-1}-(bc)^{-1}&=2(bc)^{-1}-2(ab)^{-1} \
(cd)^{-1}+(ad)^{-1}+(ab)^{-1}&=3(bc)^{-1} \
(abcd)[(cd)^{-1}+(ad)^{-1}+(ab)^{-1}]&=(abcd)[3(bc)^{-1}] \
ab+bc+cd&=3ad \
end{align}
answered Sep 9 '18 at 4:30
YutaYuta
62929
$endgroup$
$begingroup$
Your solution is very similar to what OP has done. I think OP wants an alternate method which is shorter.
$endgroup$
– Anurag A
Sep 9 '18 at 4:39
add a comment |
$begingroup$
Since $a$, $b$, $c$, $d$ are in HP, $c^{-1}+a^{-1}=2b^{-1}$ and $d^{-1}-b^{-1}=c^{-1}-a^{-1}$.
begin{align}
(c^{-1}+a^{-1})(d^{-1}-b^{-1})&=2b^{-1}(c^{-1}-a^{-1}) \
(cd)^{-1}+(ad)^{-1}-(ab)^{-1}-(bc)^{-1}&=2(bc)^{-1}-2(ab)^{-1} \
(cd)^{-1}+(ad)^{-1}+(ab)^{-1}&=3(bc)^{-1} \
(abcd)[(cd)^{-1}+(ad)^{-1}+(ab)^{-1}]&=(abcd)[3(bc)^{-1}] \
ab+bc+cd&=3ad \
end{align}
answered Sep 9 '18 at 4:30
YutaYuta
62929
$endgroup$
$begingroup$
Your solution is very similar to what OP has done. I think OP wants an alternate method which is shorter.
$endgroup$
– Anurag A
Sep 9 '18 at 4:39
add a comment |
$begingroup$
Since $a$, $b$, $c$, $d$ are in HP, $c^{-1}+a^{-1}=2b^{-1}$ and $d^{-1}-b^{-1}=c^{-1}-a^{-1}$.
begin{align}
(c^{-1}+a^{-1})(d^{-1}-b^{-1})&=2b^{-1}(c^{-1}-a^{-1}) \
(cd)^{-1}+(ad)^{-1}-(ab)^{-1}-(bc)^{-1}&=2(bc)^{-1}-2(ab)^{-1} \
(cd)^{-1}+(ad)^{-1}+(ab)^{-1}&=3(bc)^{-1} \
(abcd)[(cd)^{-1}+(ad)^{-1}+(ab)^{-1}]&=(abcd)[3(bc)^{-1}] \
ab+bc+cd&=3ad \
end{align}
answered Sep 9 '18 at 4:30
YutaYuta
62929
$endgroup$
Since $a$, $b$, $c$, $d$ are in HP, $c^{-1}+a^{-1}=2b^{-1}$ and $d^{-1}-b^{-1}=c^{-1}-a^{-1}$.
begin{align}
(c^{-1}+a^{-1})(d^{-1}-b^{-1})&=2b^{-1}(c^{-1}-a^{-1}) \
(cd)^{-1}+(ad)^{-1}-(ab)^{-1}-(bc)^{-1}&=2(bc)^{-1}-2(ab)^{-1} \
(cd)^{-1}+(ad)^{-1}+(ab)^{-1}&=3(bc)^{-1} \
(abcd)[(cd)^{-1}+(ad)^{-1}+(ab)^{-1}]&=(abcd)[3(bc)^{-1}] \
ab+bc+cd&=3ad \
end{align}
answered Sep 9 '18 at 4:30
YutaYuta
62929
answered Sep 9 '18 at 4:30
YutaYuta
62929
answered Sep 9 '18 at 4:30
YutaYuta
62929
answered Sep 9 '18 at 4:30
YutaYuta
62929
62929
$begingroup$
Your solution is very similar to what OP has done. I think OP wants an alternate method which is shorter.
$endgroup$
– Anurag A
Sep 9 '18 at 4:39
add a comment |
$begingroup$
Your solution is very similar to what OP has done. I think OP wants an alternate method which is shorter.
$endgroup$
– Anurag A
Sep 9 '18 at 4:39
$begingroup$
Your solution is very similar to what OP has done. I think OP wants an alternate method which is shorter.
$endgroup$
– Anurag A
Sep 9 '18 at 4:39
$begingroup$
Your solution is very similar to what OP has done. I think OP wants an alternate method which is shorter.
$endgroup$
– Anurag A
Sep 9 '18 at 4:39
add a comment |
$begingroup$
To organize your method:
$$k=frac13left(frac1d-frac1aright)=frac12left(frac1c-frac1aright)=frac1b-frac1a,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$.
Equating $(1,3),(2,3),(1,2)$ we get:
$$ab=3ad-2bd\
bc=2ac-ab\
cd=3ad-2ac\
ab+bc+cd=3ad-2bd+2ac-ab+3ad-2ac=\
=3ad+(3ad-2bd)-ab=3ad.$$
Alternative method:
$$k=frac1b-frac1a=frac1c-frac1b=frac1d-frac1c,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$. It results in:
$$ab=frac{a-b}{k}; bc=frac{b-c}{k}; cd=frac{c-d}{k};\
ab+bc+cd=frac{a-b+b-c+c-d}{k}=frac{a-d}{k}=3ad,$$
because:
$$frac{a-d}{k}=3ad iff k=frac13left(frac1d-frac1aright).$$
answered Sep 9 '18 at 6:54
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
To organize your method:
$$k=frac13left(frac1d-frac1aright)=frac12left(frac1c-frac1aright)=frac1b-frac1a,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$.
Equating $(1,3),(2,3),(1,2)$ we get:
$$ab=3ad-2bd\
bc=2ac-ab\
cd=3ad-2ac\
ab+bc+cd=3ad-2bd+2ac-ab+3ad-2ac=\
=3ad+(3ad-2bd)-ab=3ad.$$
Alternative method:
$$k=frac1b-frac1a=frac1c-frac1b=frac1d-frac1c,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$. It results in:
$$ab=frac{a-b}{k}; bc=frac{b-c}{k}; cd=frac{c-d}{k};\
ab+bc+cd=frac{a-b+b-c+c-d}{k}=frac{a-d}{k}=3ad,$$
because:
$$frac{a-d}{k}=3ad iff k=frac13left(frac1d-frac1aright).$$
answered Sep 9 '18 at 6:54
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
To organize your method:
$$k=frac13left(frac1d-frac1aright)=frac12left(frac1c-frac1aright)=frac1b-frac1a,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$.
Equating $(1,3),(2,3),(1,2)$ we get:
$$ab=3ad-2bd\
bc=2ac-ab\
cd=3ad-2ac\
ab+bc+cd=3ad-2bd+2ac-ab+3ad-2ac=\
=3ad+(3ad-2bd)-ab=3ad.$$
Alternative method:
$$k=frac1b-frac1a=frac1c-frac1b=frac1d-frac1c,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$. It results in:
$$ab=frac{a-b}{k}; bc=frac{b-c}{k}; cd=frac{c-d}{k};\
ab+bc+cd=frac{a-b+b-c+c-d}{k}=frac{a-d}{k}=3ad,$$
because:
$$frac{a-d}{k}=3ad iff k=frac13left(frac1d-frac1aright).$$
answered Sep 9 '18 at 6:54
farruhotafarruhota
21.8k2842
$endgroup$
To organize your method:
$$k=frac13left(frac1d-frac1aright)=frac12left(frac1c-frac1aright)=frac1b-frac1a,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$.
Equating $(1,3),(2,3),(1,2)$ we get:
$$ab=3ad-2bd\
bc=2ac-ab\
cd=3ad-2ac\
ab+bc+cd=3ad-2bd+2ac-ab+3ad-2ac=\
=3ad+(3ad-2bd)-ab=3ad.$$
Alternative method:
$$k=frac1b-frac1a=frac1c-frac1b=frac1d-frac1c,$$
where $k$ is the common difference of the AP: $frac1a,frac1b,frac1c,frac1d$. It results in:
$$ab=frac{a-b}{k}; bc=frac{b-c}{k}; cd=frac{c-d}{k};\
ab+bc+cd=frac{a-b+b-c+c-d}{k}=frac{a-d}{k}=3ad,$$
because:
$$frac{a-d}{k}=3ad iff k=frac13left(frac1d-frac1aright).$$
answered Sep 9 '18 at 6:54
farruhotafarruhota
21.8k2842
answered Sep 9 '18 at 6:54
farruhotafarruhota
21.8k2842
answered Sep 9 '18 at 6:54
farruhotafarruhota
21.8k2842
answered Sep 9 '18 at 6:54
farruhotafarruhota
21.8k2842
21.8k2842
add a comment |
add a comment |
$begingroup$
This is no quicker, but for what it’s worth:
If $a,b,c,d$ are in harmonic progression, then $frac1a,frac1b,frac1c, frac1d$ are in arithmetic progression, so $frac1a+frac1d=frac1b+frac1c$, or (just adding the fractions) $$displaystyle frac{a+d}{color{red}{ad}}=frac{b+c}{bc}.$$
Also, $frac1a+frac1c=frac2b$ and $frac1b+frac1d=frac2c$, or $frac{a+c}{ac}=frac2b$ and $frac{b+d}{bd}=frac2c$, from which $$displaystylefrac{2a+2c}{color{green}{abc}}=frac4{b^2}mbox{ and }displaystylefrac{2b+2d}{color{blue}{bcd}}=frac4{c^2}$$.
From this, $displaystyle color{red}{ad}=frac{(a+d)bc}{b+c}=frac{abc+bcd}{b+c}$, and you therefore need to show that $displaystyle ab+bc+cd=3frac{abc+bcd}{b+c}$, or equivalently that $$displaystyle (b+c)(ab+bc+cd)=3(abc+bcd).$$
Now, $3(abc+bcd)\=(abc+bcd)+2(color{green}{abc}+color{blue}{bcd})\=abc+bcd+(ab^2+b^2c)+(bc^2+c^2d)\=ab^2+b^2c+bcd+abc+bc^2+c^2d\=b(ab+bc+cd)+c(ab+bc+cd).$
edited Jan 29 at 7:23
user69284
17918
answered Sep 9 '18 at 19:23
Steve KassSteve Kass
11.4k11530
$endgroup$
add a comment |
$begingroup$
This is no quicker, but for what it’s worth:
If $a,b,c,d$ are in harmonic progression, then $frac1a,frac1b,frac1c, frac1d$ are in arithmetic progression, so $frac1a+frac1d=frac1b+frac1c$, or (just adding the fractions) $$displaystyle frac{a+d}{color{red}{ad}}=frac{b+c}{bc}.$$
Also, $frac1a+frac1c=frac2b$ and $frac1b+frac1d=frac2c$, or $frac{a+c}{ac}=frac2b$ and $frac{b+d}{bd}=frac2c$, from which $$displaystylefrac{2a+2c}{color{green}{abc}}=frac4{b^2}mbox{ and }displaystylefrac{2b+2d}{color{blue}{bcd}}=frac4{c^2}$$.
From this, $displaystyle color{red}{ad}=frac{(a+d)bc}{b+c}=frac{abc+bcd}{b+c}$, and you therefore need to show that $displaystyle ab+bc+cd=3frac{abc+bcd}{b+c}$, or equivalently that $$displaystyle (b+c)(ab+bc+cd)=3(abc+bcd).$$
Now, $3(abc+bcd)\=(abc+bcd)+2(color{green}{abc}+color{blue}{bcd})\=abc+bcd+(ab^2+b^2c)+(bc^2+c^2d)\=ab^2+b^2c+bcd+abc+bc^2+c^2d\=b(ab+bc+cd)+c(ab+bc+cd).$
edited Jan 29 at 7:23
user69284
17918
answered Sep 9 '18 at 19:23
Steve KassSteve Kass
11.4k11530
$endgroup$
add a comment |
$begingroup$
This is no quicker, but for what it’s worth:
If $a,b,c,d$ are in harmonic progression, then $frac1a,frac1b,frac1c, frac1d$ are in arithmetic progression, so $frac1a+frac1d=frac1b+frac1c$, or (just adding the fractions) $$displaystyle frac{a+d}{color{red}{ad}}=frac{b+c}{bc}.$$
Also, $frac1a+frac1c=frac2b$ and $frac1b+frac1d=frac2c$, or $frac{a+c}{ac}=frac2b$ and $frac{b+d}{bd}=frac2c$, from which $$displaystylefrac{2a+2c}{color{green}{abc}}=frac4{b^2}mbox{ and }displaystylefrac{2b+2d}{color{blue}{bcd}}=frac4{c^2}$$.
From this, $displaystyle color{red}{ad}=frac{(a+d)bc}{b+c}=frac{abc+bcd}{b+c}$, and you therefore need to show that $displaystyle ab+bc+cd=3frac{abc+bcd}{b+c}$, or equivalently that $$displaystyle (b+c)(ab+bc+cd)=3(abc+bcd).$$
Now, $3(abc+bcd)\=(abc+bcd)+2(color{green}{abc}+color{blue}{bcd})\=abc+bcd+(ab^2+b^2c)+(bc^2+c^2d)\=ab^2+b^2c+bcd+abc+bc^2+c^2d\=b(ab+bc+cd)+c(ab+bc+cd).$
edited Jan 29 at 7:23
user69284
17918
answered Sep 9 '18 at 19:23
Steve KassSteve Kass
11.4k11530
$endgroup$
This is no quicker, but for what it’s worth:
If $a,b,c,d$ are in harmonic progression, then $frac1a,frac1b,frac1c, frac1d$ are in arithmetic progression, so $frac1a+frac1d=frac1b+frac1c$, or (just adding the fractions) $$displaystyle frac{a+d}{color{red}{ad}}=frac{b+c}{bc}.$$
Also, $frac1a+frac1c=frac2b$ and $frac1b+frac1d=frac2c$, or $frac{a+c}{ac}=frac2b$ and $frac{b+d}{bd}=frac2c$, from which $$displaystylefrac{2a+2c}{color{green}{abc}}=frac4{b^2}mbox{ and }displaystylefrac{2b+2d}{color{blue}{bcd}}=frac4{c^2}$$.
From this, $displaystyle color{red}{ad}=frac{(a+d)bc}{b+c}=frac{abc+bcd}{b+c}$, and you therefore need to show that $displaystyle ab+bc+cd=3frac{abc+bcd}{b+c}$, or equivalently that $$displaystyle (b+c)(ab+bc+cd)=3(abc+bcd).$$
Now, $3(abc+bcd)\=(abc+bcd)+2(color{green}{abc}+color{blue}{bcd})\=abc+bcd+(ab^2+b^2c)+(bc^2+c^2d)\=ab^2+b^2c+bcd+abc+bc^2+c^2d\=b(ab+bc+cd)+c(ab+bc+cd).$
edited Jan 29 at 7:23
user69284
17918
answered Sep 9 '18 at 19:23
Steve KassSteve Kass
11.4k11530
edited Jan 29 at 7:23
user69284
17918
edited Jan 29 at 7:23
user69284
17918
edited Jan 29 at 7:23
user69284
17918
17918
answered Sep 9 '18 at 19:23
Steve KassSteve Kass
11.4k11530
answered Sep 9 '18 at 19:23
Steve KassSteve Kass
11.4k11530
answered Sep 9 '18 at 19:23
Steve KassSteve Kass
11.4k11530
11.4k11530
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$begingroup$
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$endgroup$
– N. F. Taussig
Sep 9 '18 at 7:24