How do you factor #8y^3+ 1#?

Answer 1

#(2y+1)(4y^2-2y+1)#

Given, #8y^3+1# #rArr (2y)^3+(1)^3# #rArr (2y+1)^3-3.2y.1(2y+1)# #rArr (2y+1)[(2y+1)^2-6y]# #rArr (2y+1)[4y^2+4y+1-6y]# #rArr (2y+1)(4y^2-2y+1)#

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Answer 2

Jameson Allen

To factor ( 8y^3 + 1 ), you can use the sum of cubes formula, which states that ( a^3 + b^3 = (a + b)(a^2 - ab + b^2) ). In this case, ( a = 2y ) and ( b = 1 ). So, ( 8y^3 + 1 = (2y + 1)(4y^2 - 2y + 1) ).

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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.

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