What is (y^3+1)-[(y^2+1)+(3y-3)] ?

The solution to (y^3+1)-[(y^2+1)+(3y-3)] is y^3-y^2-3y+3

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simplify (y^3+1)-[(y^2+1)+(3y-3)]

simplify

(y^{3} + 1) - [(y^{2} + 1) + (3 y - 3)]

y^{3} - y^{2} - 3 y + 3

Solution steps

(y^{3} + 1) - ((y^{2} + 1) + (3 y - 3))

Apply rule:

(a) = a

(y^{3} + 1) = y^{3} + 1

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

Simplify

- ((y^{2} + 1) + (3 y - 3)) : - y^{2} - 3 y + 2

= y^{3} + 1 - y^{2} - 3 y + 2

Group like terms

= y^{3} - y^{2} - 3 y + 1 + 2

Add the numbers:

1 + 2 = 3

= y^{3} - y^{2} - 3 y + 3

\frac{18}{33}

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

- ∣ 6 ∣

- 4160 - 290

- 4 - \frac{1}{3}

What is (y^3+1)-[(y^2+1)+(3y-3)] ?
The solution to (y^3+1)-[(y^2+1)+(3y-3)] is y^3-y^2-3y+3