Answer:
√x³=√(x²·x)=x√x
√xy³z=√(x·y²·y·z)=y√xyz
√18x²y=√(3²·2·x²·y)=3x√2y
√32x⁴y²=√(2⁴·2·x⁴·y²)=2²x²y√2=4x²y√2
2√16x⁷y³z⁴=2√(2⁴·x⁶·x·y²·y·z⁴)=2³x³yz²√xy=8x³yz²√xy
-3√180x⁵y=-3√(3²·2²·5·x⁴·x·y)=-3(3)(2)x²√5xy=-18x²√5xy
5√108xyz²=5√(2²·3²·3·x·y·z²)=5(2)(3)z√3xy=30z√3xy
3x√xy¹⁰z⁷=3x√x·y¹⁰·z⁶·z)=3xy⁵z³√xz
x²√x³y²=x²√(x²·x·y²)=x²·x·y√x=x³y√x
The answer is no (all you do it plug in the point that correspond with the x and y, in the equation
