Question

If a + b + c = 16 and a2 + b2 + c2 = 90, then find the

value of a3 + b3 + c3 – 3abc.​

Answer 1

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Answer 2

$Given \: a + b + c = 16 \: --(1) \\and \: a^{2} + b^{2} + c^{2} = 90 \: --(2)$

/* By Algebraic Identity */

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

$\implies 90 + 2(ab+bc+ca) = 16^{2}$

$\implies 2(ab+bc+ca) = 256 - 90$

$\implies 2(ab+bc+ca) = 166$

$\implies ab+bc+ca= \frac{166}{2}$

$\implies ab+bc+ca= 83\: --(3)$

$Now , \red { Value \:of \: a^{3} + b^{3} +c^{3} -3abc } \\= (a+b+c)(a^{2}+b^{2}+c^{2} - ab-bc-ca ) \\= (a+b+c)[a^{2}+b^{2}+c^{2} - (ab+bc+ca )]\\= 16 ( 90 - 83 ) \\= 16 \times 7 \\= 112$

Therefore.,

$\red { Value \:of \: a^{3} + b^{3} +c^{3} -3abc } \\\green { = 112 }$

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