What is 6(t+1)^3-2(t+1)^2-4(t+1)+1 ?

The solution to 6(t+1)^3-2(t+1)^2-4(t+1)+1 is 6t^3+16t^2+10t+1

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simplify 6(t+1)^3-2(t+1)^2-4(t+1)+1

simplify

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

6 t^{3} + 16 t^{2} + 10 t + 1

Solution steps

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

Expand

6 (t + 1)^{3} : 6 t^{3} + 18 t^{2} + 18 t + 6

= 6 t^{3} + 18 t^{2} + 18 t + 6 - 2 (t + 1)^{2} - 4 (t + 1) + 1

Expand

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

= 6 t^{3} + 18 t^{2} + 18 t + 6 - 2 t^{2} - 4 t - 2 - 4 (t + 1) + 1

Expand

- 4 (t + 1) : - 4 t - 4

= 6 t^{3} + 18 t^{2} + 18 t + 6 - 2 t^{2} - 4 t - 2 - 4 t - 4 + 1

Group like terms

= 6 t^{3} + 18 t^{2} - 2 t^{2} + 18 t - 4 t - 4 t + 6 - 2 - 4 + 1

Add similar elements:

18 t^{2} - 2 t^{2} = 16 t^{2}

= 6 t^{3} + 16 t^{2} + 18 t - 4 t - 4 t + 6 - 2 - 4 + 1

Add similar elements:

18 t - 4 t - 4 t = 10 t

= 6 t^{3} + 16 t^{2} + 10 t + 6 - 2 - 4 + 1

Add the numbers:

6 - 2 - 4 + 1 = 1

= 6 t^{3} + 16 t^{2} + 10 t + 1

(x + y + z)^{2} + (x + \frac{y}{2} + \frac{z}{3})^{2} - (\frac{x}{2} + \frac{y}{3} + \frac{z}{4})^{2}

x^{2} + 6 x - y^{2} + 9

2 x^{5} + 432 x^{2} y^{3}

12 x^{2} y + 6 x y^{3}

(\frac{m ^{k + 1} - 1}{2}) (\frac{m ^{k + 1} + 1}{4})

What is 6(t+1)^3-2(t+1)^2-4(t+1)+1 ?
The solution to 6(t+1)^3-2(t+1)^2-4(t+1)+1 is 6t^3+16t^2+10t+1