If a+b=1, find the value of a^{3}+b^{3}-ab-(a | vedclass

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(C) Given $a+b=1$.We need to find the value of $E = a^{3}+b^{3}-ab-(a^{2}-b^{2})^{2}$.Using the identity $a^{3}+b^{3} = (a+b)(a^{2}-ab+b^{2})$ and $a^

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(C) Given $a+b=1$.We need to find the value of $E = a^{3}+b^{3}-ab-(a^{2}-b^{2})^{2}$.Using the identity $a^{3}+b^{3} = (a+b)(a^{2}-ab+b^{2})$ and $a^{2}-b^{2} = (a+b)(a-b)$:$E = (a+b)(a^{2}-ab+b^{2}) - ab - [(a+b)(a-b)]^{2}$.Since $a+b=1$,substitute $1$ for $(a+b)$:$E = 1 \cdot (a^{2}-ab+b^{2}) - ab - (1)^{2}(a-b)^{2}$.$E = a^{2}-ab+b^{2} - ab - (a^{2}-2ab+b^{2})$.$E = a^{2}-2ab+b^{2} - a^{2}+2ab-b^{2}$.$E = 0$.

If $a+b=1,$ find the value of $a^{3}+b^{3}-ab-(a^{2}-b^{2})^{2}$.

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