Answer:
= x⁵z
Explanation:
Expression given in the question:
![((x^2y)^2(xy)^3z^2))/(((xy^2)^2yz))](https://img.qammunity.org/2020/formulas/mathematics/high-school/fb4evkpxdkmsqk5kdvpj29hbyuvws918bt.png)
now,
when the power is applied to the number with power, the power of the number gets multiplied i.e
(Xᵃ)ᵇ = Xᵃᵇ
The number having same base when multiplied together, the powers of the numbers gets added
Xᵃ × Xᵇ = Xᵃ⁺ᵇ
and,
The number having same base are when divided , the powers of the numbers gets subtracted
= Xᵃ⁻ᵇ
thus using the above property, we get
⇒
![\frac{(x^(2)*2}y^2)(x^3y^3)z^2)}{((x^2y^(2*2))yz)}](https://img.qammunity.org/2020/formulas/mathematics/high-school/54qr3ufz0pxunptdyx65cglu2obpdeaulo.png)
or
⇒
![((x^(4)y^2)(x^3y^3)z^2))/(((x^2y^(4))yz))](https://img.qammunity.org/2020/formulas/mathematics/high-school/bo9r6kx9v378993r8ya2kutc53279zfovk.png)
or
⇒
![((x^(4)x^3y^2y^3)z^2))/((x^2y^(4)yz))](https://img.qammunity.org/2020/formulas/mathematics/high-school/j3p4q9a07b5243oxy9nzfe8bdzykf3yxus.png)
or
⇒
![((x^(4+3)y^(2+3))z^2))/((x^2y^(4+1)z))](https://img.qammunity.org/2020/formulas/mathematics/high-school/oaquaqmv2j5f5ld2ag1cfibcoilnktvfy0.png)
or
⇒
![((x^(7)y^(5))z^2))/((x^2y^(5)z))](https://img.qammunity.org/2020/formulas/mathematics/high-school/lgyvvmbah8k1fw9oavmv6tcl2s0v8yldok.png)
or
⇒
![(x^(7-2)y^(5-5))z^(2-1))](https://img.qammunity.org/2020/formulas/mathematics/high-school/f7rc4rlflxysfiowoonvpd1o6cepycyxu8.png)
or
⇒
![(x^(5)y^(0))z^(1))](https://img.qammunity.org/2020/formulas/mathematics/high-school/eq852u0jo3620ii0vxa0w6s1tlo91idlef.png)
or
⇒ x⁵z