Question

Please look at the question and solve! Thanks!

Answer 1

Answer:

  $x^{27}y^{24}z^{51}$

Step-by-step explanation:

$\left(\dfrac{(x^3yz^4)^2(xy^2z^3)^4}{xy^2z^3}\right)^3=x^{3(3\cdot 2+1\cdot 4-1)}y^{3(1\cdot 2+2\cdot 4-2)}z^{3(4\cdot 2+3\cdot 4-3)}\\\\=x^{3(6+4-1)}y^{3(2+8-2)}z^{3(8+12-3)}=x^{27}y^{24}z^{51}$

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You can recognize that the denominator factor is also one  of the numerator factors, so can be cancelled right away. Proceeding to evaluate inside parentheses first, we get ...

$\left(\dfrac{(x^3yz^4)^2(xy^2z^3)^4}{xy^2z^3}\right)^3=\left((x^3yz^4)^2(xy^2z^3)^3\right)^3\\\\=(x^6y^2z^8x^3y^6z^9)^3=(x^9y^8z^{17})^3\\\\=x^{27}y^{24}z^{51}$

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The rules of exponents that apply are ...

$(a^b)(a^c)=a^{b+c}\\\\(a^b)^c=a^{b\cdot c}$