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

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

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

simplify

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

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

Solution steps

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

Apply rule:

(a) = a

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

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

Simplify

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

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

Group like terms

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

Add the numbers:

1 + 1 = 2

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

- 5 x + 6 x

x^{3} y - 81 x y

e^{l n (x^{4})}

3^{m} \cdot 2 7^{m} \cdot 9^{m}

(- 3 x) + (2 x^{2} - 8 x)

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