What is expand (d^3-2d)^4 ?

The solution to expand (d^3-2d)^4 is d^{12}-8d^{10}+24d^8-32d^6+16d^4

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expand (d^3-2d)^4

expand

(d^{3} - 2 d)^{4}

d^{12} - 8 d^{10} + 24 d^{8} - 32 d^{6} + 16 d^{4}

Solution steps

(d^{3} - 2 d)^{4}

Apply binomial theorem:

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

a = d^{3}, b = - 2 d

= i = 0 \sum 4 (i 4) (d^{3})^{(4 - i)} (- 2 d)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (d^{3})^{4} (- 2 d)^{0} : d^{12}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (d^{3})^{3} (- 2 d)^{1} : - 8 d^{10}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (d^{3})^{2} (- 2 d)^{2} : 24 d^{8}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (d^{3})^{1} (- 2 d)^{3} : - 32 d^{6}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (d^{3})^{0} (- 2 d)^{4} : 16 d^{4}

= d^{12} - 8 d^{10} + 24 d^{8} - 32 d^{6} + 16 d^{4}

(4 - d)^{3}

\frac{( - 3 x ^{5} y ^{3} )}{( - 4 x ^{2} y )}

sin^{2} (x) sec^{2} (x)

\frac{5 y ^{2} - 20}{y ^{2} - 9 y + 14}

(b + c)^{3}

What is expand (d^3-2d)^4 ?
The solution to expand (d^3-2d)^4 is d^{12}-8d^{10}+24d^8-32d^6+16d^4