Question

please help me please yaar please ​

Answer 1

Required answer: $56$ and $27$ respectively.

Solution

Since the value is $a^3+b^3+c^3-3abc$, it can be factorized into $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

We find either $a+b+c$ or $a^2+b^2+c^2$ or $ab+bc+ca$ to find its value.

(I) $a+b+c=14,\ a^2+b^2+c^2=68$

$\implies (a+b+c)^2-(a^2+b^2+c^2)=14^2-68$

$\implies 2ab+2bc+2ca=128$

$\implies ab+bc+ca=64$

The value to be found:-

$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$=14\times (68-64)$

$=14\times 4=\boxed{56}$

(II) $a+b+c=9,\ ab+bc+ca=26$

$\implies (a+b+c)^2-2(ab+bc+ca)=81-52$

$\implies a^2+b^2+c^2=29$

The value to be found:-

$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$=9\times (29-26)$

$=9\times 3=\boxed{27}$