What is expand (5x^2-3x)^4 ?

The solution to expand (5x^2-3x)^4 is 625x^8-1500x^7+1350x^6-540x^5+81x^4

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expand (5x^2-3x)^4

expand

(5 x^{2} - 3 x)^{4}

625 x^{8} - 1500 x^{7} + 1350 x^{6} - 540 x^{5} + 81 x^{4}

Solution steps

(5 x^{2} - 3 x)^{4}

Apply binomial theorem:

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

a = 5 x^{2}, b = - 3 x

= i = 0 \sum 4 (i 4) (5 x^{2})^{(4 - i)} (- 3 x)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (5 x^{2})^{4} (- 3 x)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (5 x^{2})^{3} (- 3 x)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (5 x^{2})^{2} (- 3 x)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (5 x^{2})^{1} (- 3 x)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (5 x^{2})^{0} (- 3 x)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (5 x^{2})^{4} (- 3 x)^{0} : 625 x^{8}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (5 x^{2})^{3} (- 3 x)^{1} : - 1500 x^{7}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (5 x^{2})^{2} (- 3 x)^{2} : 1350 x^{6}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (5 x^{2})^{1} (- 3 x)^{3} : - 540 x^{5}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (5 x^{2})^{0} (- 3 x)^{4} : 81 x^{4}

= 625 x^{8} - 1500 x^{7} + 1350 x^{6} - 540 x^{5} + 81 x^{4}

l^{- 1} \cdot {lo g_{10} (\frac{( s ^{2} - a ^{2} )}{( s ^{2} )})}

\frac{( - 8 x )}{( - 8 x )}

2 lo g_{10} (n^{n})

(3 p - 4 q)^{3}

\frac{101}{2} \times \frac{13}{7} \times \frac{42}{5}

What is expand (5x^2-3x)^4 ?
The solution to expand (5x^2-3x)^4 is 625x^8-1500x^7+1350x^6-540x^5+81x^4