What is expand (2x-1)^4-cos(x^2)-x^{5/3}+c ?

The solution to expand (2x-1)^4-cos(x^2)-x^{5/3}+c is 16x^4-32x^3+24x^2-8x+1-cos(x^2)-x^{5/3}+c

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expand (2x-1)^4-cos(x^2)-x^{5/3}+c

expand

(2 x - 1)^{4} - cos (x^{2}) - x^{\frac{5}{3}} + c

16 x^{4} - 32 x^{3} + 24 x^{2} - 8 x + 1 - cos (x^{2}) - x^{\frac{5}{3}} + c

Solution steps

(2 x - 1)^{4} - cos (x^{2}) - x^{\frac{5}{3}} + c

Apply binomial theorem:

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

a = 2 x, b = - 1

= i = 0 \sum 4 (i 4) (2 x)^{(4 - i)} (- 1)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (2 x)^{4} (- 1)^{0} : 16 x^{4}

Simplify

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

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (2 x)^{2} (- 1)^{2} : 24 x^{2}

Simplify

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

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (2 x)^{0} (- 1)^{4} : 1

= 16 x^{4} - 32 x^{3} + 24 x^{2} - 8 x + 1 - cos (x^{2}) - x^{\frac{5}{3}} + c

(x + 5 i - 2) (x + 2 + 5 i) (x - 5)^{2}

\frac{ln ( ( y ^{2} - 1 ) ( x ^{2} - 1 ) )}{2}

(cos (θ) + i sin (θ))^{6}

\frac{d}{d v} ((p) (p + \frac{9.5}{v ^{2}}) (v - 9.8))

3 (a - \frac{1}{6})^{2} - \frac{49}{12}

What is expand (2x-1)^4-cos(x^2)-x^{5/3}+c ?
The solution to expand (2x-1)^4-cos(x^2)-x^{5/3}+c is 16x^4-32x^3+24x^2-8x+1-cos(x^2)-x^{5/3}+c