Question

If a + b + c = 9 and a^2 + b^2 + c^2 = 35, then find the value of a^3 + b^3 + c^3 – 3abc.

Answer 1

ANSWER IN ATTACHMENT:

Answer 2

GIVEN:-

• If a+b+c = 9 and a²+b²+c²= 35.

TO FIND:-

• a³+b³+c³-3abc.

IDENTITIES USED:-

• ${\boxed{\rm{(a+b+c)^2}}}$

• ${\boxed{\rm{a^3+b^3+c^3-3abc}}}$.

Now,

• First we have to find the value of ab+bc+ac.

$\implies\rm{(a+b+c)^2=a^2+b^2+2(ab+bc+ac)}$

$\implies\rm{(9)^2=35+2(ab+bc+ac)}$

$\implies\rm{81-35=2(ab+bc+ac)}$

$\implies\rm{46=2(ab+bc+ac)}$

$\implies\rm{ab+bc+ac=23}$.

Now again,

$\implies\rm{a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)}$

$\implies\rm{a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+ac+bc)}$

$\implies\rm{a^3+b^3+c^3-3abc=(9)(35-23)}$

$\implies\rm{a^3+b^3+c^3-3abc=9\times{12}}$

$\implies\rm{a^3+b^3+c^3-3abc=108}$

Hence, The value is 108.

$\boxed{\begin{minipage}{8cm} .....\bf{some useful identities}\\ \\(a+b)^2=a^2+b^2+2ab\\ \\(a-b)^2=a^2-2ab+b^2\\ \\a^2-b^2=(a+b)(a-b)\end{minipage}}$