Answer:
The factorization of [tex]x^{3}+8[/tex] is [tex](x+2)(x^{2} -2x+4)[/tex]
Step-by-step explanation:
The problem is a sum of cubes factorization, this type of factorization applies only in binomials of the form [tex](a^{3} +b^{3} )[/tex] which means numbers that have exact cubic root and the exponents of the letters a and b are multiples of three.
Sum of cubes equation
[tex](a^{3} +b^{3} )= (a+b)(a^{2} -ab+b^{2})[/tex]
So, let's factor [tex]x^{3}+8[/tex]
we have to bring the equation to the form [tex](a^{3} +b^{3} )[/tex]
[tex]x^{3}+8=x^{3}+2^{3}[/tex] con [tex]a=x[/tex] y [tex]b=2[/tex]
Solving using sum of cubes equation
[tex](x^{3} +2^{3} )= (x+2)(x^{2} -(x)(2)+2^{2})[/tex]
[tex](x^{3} +2^{3} )=(x+2)(x^{2} -2x+4)[/tex]
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