Answer:
The answer to your question is the first option.
2x²y³ [tex]\sqrt[3]{2x^{2}y}[/tex]
Step-by-step explanation:
[tex]\sqrt[3]{32x^{8}y^{10} }[/tex]
Process
1.- Find the prime factors of 32
32 2
16 2
8 2
4 2
1
32 = 2 x 2 x 2 x2 = 2⁴
2.- Express as fractional exponents
= [tex]\sqrt[3]{2^{4} x^{8} y^{10} }[/tex]
= [tex]2^{4/3} x^{8/3} y^{10/3}[/tex]
3.- Simplify using exponent rules
= [tex]2^{3/3} 2^{1/3} x^{6/3} x^{2/3} y^{9/3} y^{1/3}[/tex]
= 2x²y³ [tex]\sqrt[3]{2x^{2}y}[/tex]
