What is expand (np-1)^4 ?

The solution to expand (np-1)^4 is n^4p^4-4n^3p^3+6n^2p^2-4np+1

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expand (np-1)^4

expand

(n p - 1)^{4}

n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + 1

Solution steps

(n p - 1)^{4}

Apply binomial theorem:

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

a = n p, b = - 1

= i = 0 \sum 4 (i 4) (n p)^{(4 - i)} (- 1)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (n p)^{4} (- 1)^{0} : n^{4} p^{4}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (n p)^{3} (- 1)^{1} : - 4 n^{3} p^{3}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (n p)^{2} (- 1)^{2} : 6 n^{2} p^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (n p)^{1} (- 1)^{3} : - 4 n p

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (n p)^{0} (- 1)^{4} : 1

= n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + 1

\frac{- 10 - 7 i}{6 + 8 i}

\frac{28}{9}

lo g_{3} (\frac{1}{27})

2 lo g_{10} (9)

(x + 2) (x - 3) (x + 4)

What is expand (np-1)^4 ?
The solution to expand (np-1)^4 is n^4p^4-4n^3p^3+6n^2p^2-4np+1