Question

if a + b + c = 0, then find the value of (a + b - c)³ + (c + a - b)³ + (b + c - a)³.

Answer 1

Explanation-

Given that, a + b + c = 0.

We have to find the value of (a + b - c)³ + (c + a - b)³ + (b + c - a)³

a + b + c = 0

We can also write it like,

a + b = - c and a + c = - b and b + c = - a

Put value of (a+b), (a+c) and (b+c) in (a+b-c)³ + (c+a-b)³ + (b+c-a)³

=> (-c - c)³ + (-b - b)³ + (-a - a)³

=> (-2c)³ + (-2b)³ + (-2a)³

=> -8c³ - 8b³ - 8a³

Take -8 as common

=> -8(c³ + b³ + a³)

=> -8(a³ + b³ + c³)

We know that a³+b³+c³ = 3abc

=> -8(3abc)

=> -24abc

•°• -24abc is the value of (a+b-c)³ + (c+a-b)³ + (b+c-a)³

Answer 2

Answer:

$\huge{\underline{\mathfrak{Answer\::}}}$

$\huge\bold\red{Given\::}$

★a + b + c = 0

$\huge\bold\red{To find\::}$

$Value\:of\:{(a+b+c)}^{3}+{(c+a-b)}^{3}+{(b+c-a)}^{3}$

Now,

a + b + c = 0

Or, a + b = - c ......(i)

Or, b + c = - a ......(ii)

Or, c + a = - b .......(iii)

Now putting the values of (i), (ii) and (iii) in the given quantity,

${( - c - c)}^{3}+{(- b - b)}^{3}+ {( - a - a)}^{3}$

${(-2c)}^{3}+{(-2b)}^{3}+{(-2a)}^{3}$

${-8c}^{3}+{-8b}^{3}+{-8a}^{3}$

Taking -8 as common

$-8({a}^{3}+{b}^{3}+{c}^{3})$

We know that,

${a}^{3}+{b}^{3}+{c}^{3}=3abc$

Then we get

-8 ( 3abc)

★= -24abc★

Hope it helps you ♥ ♥ ♥