Math, asked by dhunganaloknath73, 9 months ago

if 2s=a+b+c prove that a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)+2(s-a)(s-b)(s-c)=abc

Answers

Answered by MaheswariS
4

\textbf{Given:}

s=\dfrac{a+b+c}{2}

\textbf{To prove:}

a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)+2\,(s-a)(s-b)(s-c)=abc

\textbf{Solution:}

\text{Consider,}

a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)+2\,(s-a)(s-b)(s-c)

=a[s^2-(b+c)s+bc]+b[s^2-(c+a)s+ac]+c[s^2-(a+b)s+ab]+2[s^3-(a+b+c)s^2+(ab+bc+ca)s-abc]

=[as^2-(ab+ac)s+abc]+[bs^2-(bc+ab)s+abc]+[cs^2-(ac+bc)s+abc]+[2s^3-2(a+b+c)s^2+2(ab+bc+ca)s-2abc]

=(a+b+c)s^2-2(ab+bc+ac)s+3\,abc+2s^3-2(a+b+c)s^2+2(ab+bc+ca)s-2abc

=(a+b+c)s^2+3\,abc+2s^3-2(a+b+c)s^2-2abc

=2s^3-(a+b+c)s^2+abc

=2(\dfrac{a+b+c}{2})^3-(a+b+c)(\dfrac{a+b+c}{2})^2+abc

=2(\dfrac{(a+b+c)^3}{8})-\dfrac{(a+b+c)^3}{4}+abc

=\dfrac{(a+b+c)^3}{4}-\dfrac{(a+b+c)^3}{4}+abc

=abc

\implies\boxed{\bf\,a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)+2\,(s-a)(s-b)(s-c)=abc}

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