If left{frac{1}{2}(a-b)right}^{2}+a b=p(a+b)^ | vedclass

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(A) Given the equation: $\left(\frac{1}{2}(a-b)\right)^{2}+a b=p(a+b)^{2}$Expanding the left side: $\frac{1}{4}(a-b)^{2}+a b=p(a+b)^{2}$Multiply the e

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$\frac{1}{4}$

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(A) Given the equation: $\left(\frac{1}{2}(a-b)ight)^{2}+a b=p(a+b)^{2}$Expanding the left side: $\frac{1}{4}(a-b)^{2}+a b=p(a+b)^{2}$Multiply the entire equation by $4$ to eliminate the fraction: $(a-b)^{2}+4 a b=4 p(a+b)^{2}$Expand $(a-b)^{2}$ as $a^{2}+b^{2}-2 a b$: $(a^{2}+b^{2}-2 a b)+4 a b=4 p(a+b)^{2}$Simplify the expression: $a^{2}+b^{2}+2 a b=4 p(a+b)^{2}$Recognize that $a^{2}+b^{2}+2 a b = (a+b)^{2}$: $(a+b)^{2}=4 p(a+b)^{2}$Since $a 
eq -b$,we have $(a+b)^{2} 
eq 0$. Dividing both sides by $(a+b)^{2}$ gives: $1=4 p$Therefore,$p = \frac{1}{4}$.

If $\left\{\frac{1}{2}(a-b)\right\}^{2}+a b=p(a+b)^{2},$ then the value of $p$ is (assume that $a \neq-b).$

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