Question

(a+b)² + (b-a)² = a² + b²
prove that

Answer 1

Answer:

Let first term be :

(a - b)(a - b)(a−b)(a−b)

=(a-b)²

second term=

{a }^{ 2} + {b}^{2}a

2

+b

2

d=second term - first term

=

{a}^{2} + {b}^{2} - ( {a}^{2} - 2ab + {b}^{2} )a

2

+b

2

−(a

2

−2ab+b

2

)

=2ab

third term= first term+2(d)

=

{a}^{2} - 2ab + {b }^{2} + 2(2ab)a

2

−2ab+b

2

+2(2ab)

= {a}^{2} + 2ab + {b}^{2}=a

2

+2ab+b

2

= {(a + b)}^{2}=(a+b)

2

Hence , it forms an A.P. :

(a-b)²,(a²+b²) & (a+b)²

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Answer 2

The result of your question seems to be wrong, Kindly recheck

(a+b) ²= a²+ 2ab+ b²

(b-a) ²= a² -2ab +b²

So, (a+b) ² + (b-a) ² = a² +2ab+ b² +a² -2ab +b²

=2(a²+b²)

Hope it helps!