What is expand (2x-3y^4z^2)^4 ?

The solution to expand (2x-3y^4z^2)^4 is 16x^4-96x^3y^4z^2+216x^2y^8z^4-216xy^{12}z^6+81y^{16}z^8

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expand (2x-3y^4z^2)^4

expand

(2 x - 3 y^{4} z^{2})^{4}

16 x^{4} - 96 x^{3} y^{4} z^{2} + 216 x^{2} y^{8} z^{4} - 216 x y^{12} z^{6} + 81 y^{16} z^{8}

Solution steps

(2 x - 3 y^{4} z^{2})^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = 2 x, b = - 3 y^{4} z^{2}

= i = 0 \sum 4 (i 4) (2 x)^{(4 - i)} (- 3 y^{4} z^{2})^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (2 x)^{4} (- 3 y^{4} z^{2})^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (2 x)^{3} (- 3 y^{4} z^{2})^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (2 x)^{2} (- 3 y^{4} z^{2})^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (2 x)^{1} (- 3 y^{4} z^{2})^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (2 x)^{0} (- 3 y^{4} z^{2})^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (2 x)^{4} (- 3 y^{4} z^{2})^{0} : 16 x^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (2 x)^{3} (- 3 y^{4} z^{2})^{1} : - 96 x^{3} y^{4} z^{2}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (2 x)^{2} (- 3 y^{4} z^{2})^{2} : 216 x^{2} y^{8} z^{4}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (2 x)^{1} (- 3 y^{4} z^{2})^{3} : - 216 x y^{12} z^{6}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (2 x)^{0} (- 3 y^{4} z^{2})^{4} : 81 y^{16} z^{8}

= 16 x^{4} - 96 x^{3} y^{4} z^{2} + 216 x^{2} y^{8} z^{4} - 216 x y^{12} z^{6} + 81 y^{16} z^{8}

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What is expand (2x-3y^4z^2)^4 ?
The solution to expand (2x-3y^4z^2)^4 is 16x^4-96x^3y^4z^2+216x^2y^8z^4-216xy^{12}z^6+81y^{16}z^8