If \((x-4)\left(x^2+4x+16\right)=x^3-P\), find the value of \(\sqrt{P}\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 8
The left side matches the difference of cubes, \((a-b)(a^2+ab+b^2)=a^3-b^3\), with \(a=x\) and \(b=4\).
\((x-4)\left(x^2+4x+4^2\right)=x^3-4^3=x^3-64\)
So \(P=64\)
\(\sqrt{P}=\sqrt{64}=8\)