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How do you factor #-49 + y^6#?

Answer 1

Isaiah Anthony

#y^6 - 49 = (y^3 - 7)(y^3 + 7)#

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Answer 2

Savannah Anderson

#y^6-49=(y^3+7)(y^3-7)#

Problem: Factor #-49+y^6#.

Rewrite the equation as #y^6-49#.

An illustration of the difference of squares is as follows:

#(a^2-b^2)=(a+b)(a-b)#.

#a=y^3# #b=7#

#(y^3)^2-7^2=(y^3+7)(y^3-7)#

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Answer 3

Nathan Axel

#-49+y^6 = y^6-49 = (y^3)^2-7^2 = (y^3 - 7)(y^3 + 7)#

With rational coefficients, you can go no further than this.

If real coefficients are permitted, then:

#y^3-7 = y^3-(root(3)(7))^3#

#= (y - root(3)(7))(y^2+(root(3)(7))y+(root(3)(7))^2)#

and

#y^3+7 = y^3+(root(3)(7))^3#

#= (y + root(3)(7))(y^2-(root(3)(7))y+(root(3)(7))^2)#

For shorthand I will write #rho = root(3)(7)#

#-49+y^6#

#= (y - rho)(y^2+rho y+rho^2)(y + rho)(y^2-rho y+rho^2)#

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Answer 4

Jameson Allen

To factor the expression -49 + y^6, we can rewrite it as (y^3)^2 - 7^2. This expression can be factored using the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b). Therefore, -49 + y^6 can be factored as (y^3 + 7)(y^3 - 7).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.