What is (2x+2)^4-(x+1)^4 ?

The solution to (2x+2)^4-(x+1)^4 is 15x^4+60x^3+90x^2+60x+15

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simplify (2x+2)^4-(x+1)^4

simplify

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

15 x^{4} + 60 x^{3} + 90 x^{2} + 60 x + 15

Solution steps

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

(2 x + 2)^{4} : 16 x^{4} + 64 x^{3} + 96 x^{2} + 64 x + 16

= 16 x^{4} + 64 x^{3} + 96 x^{2} + 64 x + 16 - (x + 1)^{4}

(x + 1)^{4} : x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1

= 16 x^{4} + 64 x^{3} + 96 x^{2} + 64 x + 16 - (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1)

- (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1) : - x^{4} - 4 x^{3} - 6 x^{2} - 4 x - 1

= 16 x^{4} + 64 x^{3} + 96 x^{2} + 64 x + 16 - x^{4} - 4 x^{3} - 6 x^{2} - 4 x - 1

Simplify

16 x^{4} + 64 x^{3} + 96 x^{2} + 64 x + 16 - x^{4} - 4 x^{3} - 6 x^{2} - 4 x - 1 : 15 x^{4} + 60 x^{3} + 90 x^{2} + 60 x + 15

= 15 x^{4} + 60 x^{3} + 90 x^{2} + 60 x + 15

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

[(a^{2} \cdot b^{3})^{3}]^{4}

f (x) - 2 x^{2} + 13 x - 15

x^{3} - y^{3} - (x - y)^{3}

f (x) + h (x) 3 x^{3} + 5 x^{2} - 2 x + 7 + x^{4} - 2 x^{3} + 8 x - 14

What is (2x+2)^4-(x+1)^4 ?
The solution to (2x+2)^4-(x+1)^4 is 15x^4+60x^3+90x^2+60x+15