Using properties of determinats, prove that

a2 2ab b2

b2 a2 2ab

2ab b2 a2

= (a3 + b3)2

Δ=a22abb2b2a22ab2abb2a2=a2+b2+2ab2abb2a2+b2+2aba22aba2+b2+2abb2a2       C1C1+C2+C3=a+b22abb2a+b2a22aba+b2b2a2=a+b212abb21a22ab1b2a2       Take common a+b2=a+b202ab-b2b2-a21a22ab0b2-a2a2-2ab      R1R1-R3 and R3R3-R2 
=-a+b2a2-2ab2ab-b2-b2-a22=-a+b22a3b-a2b2-4a2b2+2ab3-b4-a4+2a2b2=a+b2a4+b4+2a3b+2ab3-3a2b2=a+b2a2+b22+2a2+b2ab-a2b2=a+b2a2+b2-ab2=a3+b32Hence proved.

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