Algberic question

• June 17th 2013, 08:22 AM
sscgeek
Algberic question
If x^a*x^b*x^c =1 then the value of a^3+b^3+c^3 is

a)9

b)abc

c) a+b+c

d) 3 abc
• June 17th 2013, 09:55 AM
ebaines
Re: Algberic question
Given:

$x^ax^bx^c = x^{(a+b+c)} = 1$

then $a+b+c = 0$, and so $c= -(a+b)$. Hence:

$a^3 + b^3 + c^3 = a^3 + b^3 - (a+b)^3 = a^3 + b^3 -(a^3 + 3a^2b + 3b^2c + b^3) = -3ab( a+b) = 3abc$
• June 17th 2013, 10:33 AM
Soroban
Re: Algberic question
Hello, sscgeek!

Quote:

$\text{Given: }\:x^ax^bx^c \:=\:1$

$\text{Find the value of: }\:a^3+b^3+c^3$

. . $(a)\;9 \qquad (b)\;abc \qquad (c)\;a+b+c \qquad (d)\;3 abc$

We have: . $x^{a+b+c} \:=\:1 \quad\Rightarrow\quad a+b+c \:=\:0$

Cube both sides: . $(a+b+c)^3 \:=\:0^3$

. . $a^3 + 3a^2b + 3ab^2 + b^3 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 + c^3 + 6abc \:=\:0$

. . $a^3 + b^3 + c^3 + (3a^2b+2ab^2 + 3abc)$
. . . . . . . . . . . . . . $+ (3b^2c+3bc^2 + 3abc) + (3a^2c+3ac^2 {\color{blue}+\: 3abc}) {\color{blue}-\: 3abc} \:=\:0$

. . $a^3+b^3+c^3 + 3ab\underbrace{(a+b+c)}_{\text{This is 0}} + \:3bc\underbrace{(a+b+c)}_{\text{This is 0}} + \:3ac\underbrace{(a+b+c)}_{\text{This is 0}} -\:3abc\:=\:0$

. . . . . . . $a^3+b^3+c^3 - 3abc \;=\;0$

. . . . . . . . . $a^3+b^3+c^3 \;=\;3abc\;\;\hdots\;\;\text{answer (d)}$