What is (y^3+1)-[(y^2+1)+(3y-7)] ?

The solution to (y^3+1)-[(y^2+1)+(3y-7)] is y^3-y^2-3y+7

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simplify (y^3+1)-[(y^2+1)+(3y-7)]

simplify

(y^{3} + 1) - [(y^{2} + 1) + (3 y - 7)]

y^{3} - y^{2} - 3 y + 7

Solution steps

(y^{3} + 1) - ((y^{2} + 1) + (3 y - 7))

Apply rule:

(a) = a

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

= y^{3} + 1 - ((y^{2} + 1) + (3 y - 7))

Simplify

- ((y^{2} + 1) + (3 y - 7)) : - y^{2} - 3 y + 6

= y^{3} + 1 - y^{2} - 3 y + 6

Group like terms

= y^{3} - y^{2} - 3 y + 1 + 6

Add the numbers:

1 + 6 = 7

= y^{3} - y^{2} - 3 y + 7

9 + (- 16)

- 5 x - 10

- 5 x + 3 x

\frac{y ^{2} + 7 y - 18}{y ^{2} - 3 y + 2}

\frac{3 x ^{3} - 3 x}{x ^{5} - x}

What is (y^3+1)-[(y^2+1)+(3y-7)] ?
The solution to (y^3+1)-[(y^2+1)+(3y-7)] is y^3-y^2-3y+7