Question

If $a+b+c=12$, $a^2+b^2+c^2=90$, find the value of $a^3+b^3+c^2-3abc$

Answer 1

Given that,

$a+b+c=12$

$a^2+b^2+c^2=90$

Then we know that,

${(a+b+c)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

${12}^2-90=2(ab+bc+ca)$

$2(ab+bc+ca)=144-90=54$

$(ab+bc+ca)=27$

Now, we know that,

$a^3+b^3+c^3-3abc$

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

$=12\times (90-27)$

$=12\times 63$

$=756$

Hence, this is the answer.