Respuesta :

[tex]\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) \\\\ a^3-b^3 = (a-b)(a^2+ab+b^2)\\\\ -------------------------------\\\\ 64x^3+27~~ \begin{cases} 64=4^3\\ 27=3^3 \end{cases}\implies 4^3x^3+3^3\implies (4x)^3+3^3 \\\\\\ (4x+3)[(4x)^2-(4x)(3)+(3)^2]\implies (4x+3)(16x^2-12x+9)[/tex]
treyvey2007ot1ih2

Using the identity a3+b3=(a+b)(a2−ab+b2) we find:

64x3+27=(4x)3+33

=(4x+3)((4x)2−(4x)⋅3+32)

=(4x+3)(16x2−12x+9)

16x2−12x+9 has no simpler factors with real coefficients.

To check this, evaluate its discriminant:

Δ(16x2−12x+9)=(−12)2−(4×16×9)

=144−576=−432

Since Δ<0 the quadratic equation 16x2−12x+9=0 has no real roots. It has two distinct complex roots.

In case you are curious, the complex factors of 16x2−12x+9 are:

(4x+3ω) and (4x+3ω2)

where ω=−12+√32i

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