Dear Student, Δ=a22abb2b2a22ab2abb2a2=a2+b2+2ab2abb2a2+b2+2aba22aba2+b2+2abb2a2 C1→C1+C2+C3=a+b22abb2a+b2a22aba+b2b2a2=a+b212abb21a22ab1b2a2 Take common a+b2=a+b202ab-b2b2-a21a22ab0b2-a2a2-2ab R1→R1-R3 and R3→R3-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 Since, (a-b)2=a2+b2-2ab=a+b2a2+b2-ab2 Since, a3+b3=(a+b)(a2+b2-ab)=a3+b32Hence proved Regards!

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Dear Student,

Δ = |\begin{matrix} a^{2} & 2 ab & b^{2} \\ b^{2} & a^{2} & 2 ab \\ 2 ab & b^{2} & a^{2} \end{matrix}| = |\begin{matrix} a^{2} + b^{2} + 2 ab & 2 ab & b^{2} \\ a^{2} + b^{2} + 2 ab & a^{2} & 2 ab \\ a^{2} + b^{2} + 2 ab & b^{2} & a^{2} \end{matrix}| (C_{1} \to C_{1} + C_{2} + C_{3}) = |\begin{matrix} {(a + b)}^{2} & 2 ab & b^{2} \\ {(a + b)}^{2} & a^{2} & 2 ab \\ {(a + b)}^{2} & b^{2} & a^{2} \end{matrix}| = {(a + b)}^{2} |\begin{matrix} 1 & 2 ab & b^{2} \\ 1 & a^{2} & 2 ab \\ 1 & b^{2} & a^{2} \end{matrix}| (Take common {(a + b)}^{2}) = {(a + b)}^{2} |\begin{matrix} 0 & 2 ab - b^{2} & b^{2} - a^{2} \\ 1 & a^{2} & 2 ab \\ 0 & b^{2} - a^{2} & a^{2} - 2 ab \end{matrix}| (R_{1} \to R_{1} - R_{3} and R_{3} \to R_{3} - R_{2})

= - {(a + b)}^{2} [(a^{2} - 2 ab) (2 ab - b^{2}) - {(b^{2} - a^{2})}^{2}] = - {(a + b)}^{2} [2 a^{3} b - a^{2} b^{2} - 4 a^{2} b^{2} + 2 {ab}^{3} - b^{4} - a^{4} + 2 a^{2} b^{2}] = {(a + b)}^{2} [a^{4} + b^{4} - 2 a^{3} b - 2 {ab}^{3} + 3 a^{2} b^{2}] = {(a + b)}^{2} [{(a^{2} + b^{2})}^{2} - 2 (a^{2} + b^{2}) ab + a^{2} b^{2}] \{S i n c e, {(a - b)}^{2} = a^{2} + b^{2} - 2 a b\} = {(a + b)}^{2} {[a^{2} + b^{2} - ab]}^{2} \{S i n c e, a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)\} = {(a^{3} + b^{3})}^{2} Hence proved .

Regards!

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