Question

Find the square root of $64(x-y{)}^4(y+x{)}^6(x-z{)}^{10}$.

• A. $8(x-y{)}^2(y+x{)}^3(x-z{)}^5)$
• B. $(x-y{)}^2(y+x{)}^3(x-z{)}^5)$
• C. $8\left|\right(x-y{)}^4(y+x{)}^3(x-z{)}^5\left)\right|$
• D. $8\left|\right(x-y{)}^2(y+x{)}^3(x-z{)}^5\left)\right|$

Answer 1

Answer: (D)

$8\left|\right(x-y{)}^2(y+x{)}^3(x-z{)}^5\left)\right|$

Given, $64(x-y{)}^4(y+x{)}^6(x-z{)}^{10}$

$\Rightarrow \sqrt{64(x-y{)}^4(y+x{)}^6(x-z{)}^{10}}$

$=$ $\sqrt{\left(8\right(x-y{)}^2(y+x{)}^3(x-z{)}^5{)}^2}$

So, the square root of $64(x-y{)}^4(y+x{)}^6(x-z{)}^{10}=8\left|\right(x-y{)}^2(y+x{)}^3(x-z{)}^5\left)\right|$