What is (2x-1)^3-(2x-1)^2-(2x-1)-1 ?

The solution to (2x-1)^3-(2x-1)^2-(2x-1)-1 is 8x^3-16x^2+8x-2

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simplify (2x-1)^3-(2x-1)^2-(2x-1)-1

simplify

(2 x - 1)^{3} - (2 x - 1)^{2} - (2 x - 1) - 1

8 x^{3} - 16 x^{2} + 8 x - 2

Solution steps

(2 x - 1)^{3} - (2 x - 1)^{2} - (2 x - 1) - 1

Apply the distributive law:

- (a - b) = - a + b

- (2 x - 1) = - 2 x + 1

= (2 x - 1)^{3} - (2 x - 1)^{2} - 2 x + 1 - 1

Expand

(2 x - 1)^{3} : 8 x^{3} - 12 x^{2} + 6 x - 1

= 8 x^{3} - 12 x^{2} + 6 x - 1 - (2 x - 1)^{2} - 2 x + 1 - 1

Expand

- (2 x - 1)^{2} : - 4 x^{2} + 4 x - 1

= 8 x^{3} - 12 x^{2} + 6 x - 1 - 4 x^{2} + 4 x - 1 - 2 x + 1 - 1

Group like terms

= 8 x^{3} - 12 x^{2} - 4 x^{2} + 6 x + 4 x - 2 x - 1 - 1 - 1 + 1

Add similar elements:

- 12 x^{2} - 4 x^{2} = - 16 x^{2}

= 8 x^{3} - 16 x^{2} + 6 x + 4 x - 2 x - 1 - 1 - 1 + 1

Add similar elements:

6 x + 4 x - 2 x = 8 x

= 8 x^{3} - 16 x^{2} + 8 x - 1 - 1 - 1 + 1

Add the numbers:

- 1 - 1 - 1 + 1 = - 2

= 8 x^{3} - 16 x^{2} + 8 x - 2

- \frac{1}{( y ( y - 1 ) )} + \frac{( y ^{2} )}{( y - 1 )}

\frac{2 x ^{4} y ^{5}}{x ^{2} y ^{3}}

((\frac{7 x}{8}) + (\frac{3 y}{4}))^{2}

\frac{4 o ^{'} 0}{9}

tan^{2} (t) - 16 cos^{2} (t)

What is (2x-1)^3-(2x-1)^2-(2x-1)-1 ?
The solution to (2x-1)^3-(2x-1)^2-(2x-1)-1 is 8x^3-16x^2+8x-2