Question

if a+b+c=4 and ab+bc+ca=6 find a³+b³+c³-3abc

Answer 1

$\large {\huge{ \color{red}{ \mathfrak{ \star \: welcome \: to \: my \: answer \: \star \star}}}}$



$\bold{ \huge{ \boxed{required \: answer}}}$


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Answer 2

$\boxed{\large{\red{\bold{a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-(ab+bc+ca)}}}}$

$\\\implies (4)(a^{2}+b^{2}+c^{2}-6)$

we have a another formula

(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

(4)²=a²+b²+c²+2(6)

a²+b²+c²=16-12

a²+b²+c²=4

$\\\implies (4)(a^{2}+b^{2}+c^{2}-6)\\\implies (4)((4)-(6))\\\\\implies (4)(-2)\\\\\boxed{\green{\implies -8}}$