Question

If a+b+c=5 and ab +bc+ca=10,then prove that a3 +b3 +c3 -3abc= -25

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• a+b+c=5 and
• ab+bc+ac=10

${\sf{\green{\underline{\large{To\:prove}}}}}$

• a³+b³+c³-3abc=-25

${\sf{\pink{\underline{\Large{Explanation}}}}}$

• we know that

(a+b+c)²=a²+b²+c²+2(ab+bc+ac)

• a+b+c=5 (both sid squareing)

=>(a+b+c)²=5²

=>a²+b²+c²+2(ab+bc+ac)=25

=>a²+b²+c²+2(10)=25

=>a²+b²+c²+20=25

=>a²+b²+c²=5

Factorize a3 +b3 +c3 -3abc

=>a3 +b3 +c3 -3abc= (a+b+c) (a²+b²+c²-ab-bc-ac)

=>a3 +b3 +c3 -3abc=(5)(5-10)

=>a3 +b3 +c3 -3abc=-25

Hence proved