Question

plz answer my question​

Answer 1

Step-by-step explanation:

please find attached

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Answer 2

Proof is given below:

To Show That

$\Sigma ab(a+b)+2abc=(a+b)(b+c)(c+a)$

LHS

$=\Sigma [ab(a+b)]+2abc$

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

$=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+abc+abc$

$=a(ab+c^2+ac+bc)+b(ab+bc+c^2+ac)$

$=(a+b)(ab+ac+c^2+bc)$

$=(a+b)[a(b+c)+c(b+c)]$

$=(a+b)(b+c)(c+a)$

= RHS                                                     (Proved)

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Q: a^2+b^2+c^2-ab-bc-ca =0 then prove that a=b=c

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