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How do you factor completely #x^3+64#?

Answer 1

Nathan Axel

#x^3+64=(x+4)(x^2-4x+16)#

Both #x^3# and #64=4^3# are perfect squares.

So we can use the sum of cubes identity:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

with #a=x# and #b=4# as follows:

#x^3+64#

#=x^3+4^3#

#=(x+4)(x^2-x(4)+4^2)#

#=(x+4)(x^2-4x+16)#

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Answer 2

Genesis Allen

To factor completely ( x^3 + 64 ), 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 = x ) and ( b = 4 ). So, ( x^3 + 64 = (x + 4)(x^2 - 4x + 16) ). Therefore, ( x^3 + 64 ) factors completely into ( (x + 4)(x^2 - 4x + 16) ).

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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.