Respuesta :
Given that the sum of two cubes can be factored by using the formula
[tex]a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]a) To verify the formula by multiplying the right side equation
[tex]\begin{gathered} (a+b)(a^2-ab+b^2) \\ =a(a^2-ab+b^2)+b(a^2-ab+b^2) \\ =a^3-a^2b+ab^2+a^2b-ab^2+b^3 \\ \text{Collect like terms} \\ =a^3-a^2b+a^2b+ab^2-ab^2+b^3 \\ \text{Simplify} \\ =a^3+b^3 \end{gathered}[/tex]Hence,
[tex](a+b)(a^2-ab+b^2)=a^3+b^3[/tex]b) To factor
[tex]8x^3+27[/tex]Using the sum of two cubes formula, i.e
[tex]a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]Factorizing the expression gives
[tex]\begin{gathered} (2x)^3+(3)^3=(2x+3)((2x)^2-(2x)(3)+(3)^2)_{} \\ (2x)^3+(3)^3=(2x+3)(4x^2-6x+9) \end{gathered}[/tex]Hence, the answer is
[tex](2x+3)(4x^2-6x+9)[/tex]c) Given that one of the factors of a³ - b³ is a- b, the quadratic factor of a³ - b³ can be deduced by applying the differences of cubes formula below
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)^{}_{}[/tex]Expanding the right side equations
[tex]\begin{gathered} (a-b)(a^2+ab+b^2)^{}_{}=a(a^2+ab+b^2)-b(a^2+ab+b^2) \\ =a^3+a^2b+ab^2-a^2b-ab^2-b^3 \\ \text{Collect like terms} \\ =a^3+a^2b-a^2b+ab^2-ab^2-b^3 \\ \text{Simplify} \\ =a^3-b^3 \end{gathered}[/tex]Hence, the quadratic factor is
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]d) To factor the expression
[tex]x^3-1[/tex]By applying the differences of cubes formula
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]Factorizing the expression gives
[tex]\begin{gathered} (x)^3-(1)^3=(x-1)(x^2+(x)(1)+1^2)^{}_{} \\ x^3-1^3=(x-1)(x^2+x+1) \end{gathered}[/tex]Hence, the answer is
[tex](x-1)(x^2+x+1)[/tex]
