Respuesta :
[tex]\bf \textit{difference of cubes}
\\\\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3
\\\\
a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad
(a-b)(a^2+ab+b^2)= a^3-b^3\\\\
-------------------------------\\\\
a^3-b^3\implies (a-b)(a^2+ab+b^2)[/tex]
Answer:
We have to find the difference of cubes of any two real numbers
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
Verification of the Result
RHS
[tex]=(a-b)(a^2+ab+b^2)\\\\=a \times (a^2+ab+b^2)-b \times (a^2+ab+b^2)\\\\\text{Using Distributive property }\\\\=a^3+a^2b+ab^2-ba^2-ab^2-b^3\\\\ \text{Cancelling like terms}\\\\=a^3-b^3[/tex]
=LHS
