Question

If a+b+c=5,ab+bc+ca=10 prove that a^3+b^3+c^3-3abc=25

Answer 1

Given,

$a + b + c = 5 \\ \\ ab + bc + ca = 10 \\ \\ (a+b+c)^{2} = a^{2} + b^{2} +c^{2} +2(ab + ba+ca) \\ \\ 5^{2} = a^{2} + b^{2} +c^{2} +2*10 \\ \\25 = a^{2} + b^{2} +c^{2} +20 \\ \\ a^{2} + b^{2} +c^{2} = 5 \\ \\ \text{We know that} \\ \\ a^{3} + b^{3} + c^{3} -3abc = (a+b+c)(a^{2} + b^{2} + c ^{2} -ab -bc -ca ) \\ \\ a^{3} + b^{3} + c^{3} -3abc = (a+b+c)[a^{2} + b^{2} + c ^{2} -(ab+bc+ca) ] \\ \\ a^{3} + b^{3} + c^{3} -3abc = (5)[5 - 10] \\ \\ a^{3} + b^{3} + c^{3} -3abc = -25 \ \ \text{Ans.}$