Question

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

Answer 1

upHey there!!

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Solution:-

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=

${a}^{2} - {b}^{2} + ( {b}^{2} - {c}^{2} ) + ( {c}^{2} - {a}^{2} ) = 0 \\ \\ {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} = 0 \\ \\0 = 0$



LHS= RHS

Answer 2

(a^2)-(b^2)+(b^2)-(c^2)+(c^2)-(a^2)=0

a^2-b^2+b^2-c^2+c^2-a^2=0

cancelling all +/-

0=0

lhs = rhs

hence proved

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