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Simplify: (x²y³z⁻³)⁷(xy⁴)

User RaGin RAj
by
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2 Answers

2 votes

Answer:


\huge \boxed{ (x^(15) y^(25))/(z^(21) ) }

Explanation:


(x^2 y^3 z^(-3))^7 (xy^4)


\sf Apply \ exponent \ rule : (a^b)^c=a^(bc)


(x^(2 * 7) y^(3 * 7) z^(-3 * 7))(xy^4)


(x^(14) y^(21) z^(-21))(xy^4)


x^(14) * y^(21) * z^(-21) * x * y^4


\sf Apply \ exponent \ rule : a^b * a^c=a^(b+c)


x^(14) * x^1 * y^(21) * y^4 * z^(-21)


x^(14+1) * y^(21+4) * z^(-21)


x^(15) * y^(25) * z^(-21)


\displaystyle \sf Apply \ exponent \ rule : a^(-b) =(1)/(a^b)


\displaystyle x^(15) * y^(25) * (1)/(z^(21) )


\displaystyle (x^(15) y^(25))/(z^(21) )

User Lucperkins
by
5.4k points
2 votes

Answer:


\huge\boxed{(x^(15)y^(25))/(z^(21)) }

Explanation:


(x^2y^3z^(-3))^7(xy^4)\\

Expanding Parenthesis


(x^(2*7 ) y^(3*7)z^(-3*7))(xy^4)\\x^(14)y^(21)z^(-21) * xy^4

Combining like terms


x^(14) * x * y^(21) * y^4 * z^(-21)

When bases are same , powers are to be added


x^(14+1) y^(21+4) x^(-21)


x^(15) y^(25) z^(-21)

Neutralizing the negative sign by shifting z to the denominator


(x^(15)y^(25))/(z^(21))

User BJ Myers
by
5.0k points