Question

if a+b+c=6 ,then find (2-a)^3 + (2-b)^3 + (2-c)^3 -3(2-a)(2-b)(2-c)

Answer 1

ANSWER:

Given:

• a + b + c = 6

To Find:

• (2 - a)^3 + (2 - b)^3 + (2 - c)^3 - 3(2-a)(2-b)(2-c)

Solution:

We need to find the value of,

$\implies(2 - a)^3 + (2 - b)^3 + (2 - c)^3 - 3(2-a)(2-b)(2-c)$

Let us assume that,

• (2 - a) = x
• (2 - b) = y
• (2 - c) = z

So,

$\implies(2 - a)^3 + (2 - b)^3 + (2 - c)^3 - 3(2-a)(2-b)(2-c)$

$\implies x^3 + y^3 + z^3 - 3xyz$

We know that,

$\implies p^3 + q^3 + r^3 - 3pqr=(p+q+r)(p^2+q^2+r^2-pq-qr-rp)$

So,

$\implies x^3 + y^3 + z^3 - 3xyz$

$\implies(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Substituting the values,

$\implies[(2-a)+(2-b)+(2-c)][(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)]$

$\implies[2-a+2-b+2-c][(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)]$

$\implies[6-(a+b+c)][(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)]$

We are given that,

$\implies a+b+c=6$

So, substituting the above value,

$\implies[6-(a+b+c)][(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)]$

$\implies[6-6][(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)]$

$\implies[0][(2-a)^2+(2-b)^2+(2-c)^2-(2-a)(2-b)-(2-b)(2-c)-(2-c)(2-a)]$

$\implies\bf 0$

Therefore,

$\implies\bf(2 - a)^3 + (2 - b)^3 + (2 - c)^3 - 3(2-a)(2-b)(2-c)=0$