How do you factor 64r^3 - 2764r3−27?

2 Answers
sente
Feb 27, 2016

Use the difference of cubes formula (and the quadratic formula if you allow for complex numbers) to find that

64r^3-27=(4r-3)(16r^2+12r+9)64r3−27=(4r−3)(16r2+12r+9)

=(4r-3)(4r+3/2-(3sqrt(3))/2i)(4r+3/2+(3sqrt(3))/2i)=(4r−3)(4r+32−3√32i)(4r+32+3√32i)

Explanation:

The difference of cubes formula states that
a^3-b^3 = (a-b)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2)
(Try multiplying the right hand side to verify this).

In this case, that gives us

64r^3-27 = (4r)^3-3^364r3−27=(4r)3−33

=(4r-3)(16r^2+12r+9)=(4r−3)(16r2+12r+9)

If we are only consider real numbers, then we are done. If we are allowing for complex numbers, then we can continue factoring using the quadratic formula on the remaining quadratic expression.

16r^2+12r+9 = 016r2+12r+9=0

<=>r = (-12+-sqrt(12^2-4(16)(9)))/(2(16))⇔r=−12±√122−4(16)(9)2(16)

=(-12+-sqrt(-432))/32=−12±√−43232

=(-12+-12sqrt(3)i)/32=−12±12√3i32

=-3/8+-(3sqrt(3))/8i=−38±3√38i

Thus, making sure to multiply by 1616 to obtain the correct coefficients, the complete factorization would be

64r^3-27 =64r3−27=
=(4r-3)(4r+3/2-(3sqrt(3))/2i)(4r+3/2+(3sqrt(3))/2i)=(4r−3)(4r+32−3√32i)(4r+32+3√32i)

Shwetank Mauria
Feb 27, 2016

Hence factors are given by (64r^3−27)=(4r-3)(16r^2+12r+9)(64r3−27)=(4r−3)(16r2+12r+9)

Explanation:

As the function 64r^3−27=(4r)^3-3^364r3−27=(4r)3−33, one could use the identity

(x^3-y^3)=(x-y)(x^2+xy+y^2)(x3−y3)=(x−y)(x2+xy+y2).

Using this, 64r^3−27=(4r-3)((4r)^2+4rxx3+3^2)64r3−27=(4r−3)((4r)2+4r×3+32)

= (4r-3)(16r^2+12r+9)(4r−3)(16r2+12r+9)

It is obvious that this cannot be factorized further, as the determinant 12^2-4xx16xx9=-432122−4×16×9=−432, is negative.

Hence factors are given by (64r^3−27)=(4r-3)(16r^2+12r+9)(64r3−27)=(4r−3)(16r2+12r+9)

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