Respuesta :
SOLUTION:
The expression is given below as
[tex]121x^4y^6-25x^2z^{12}[/tex]Step 1:
Factor out the common term
The common term is
[tex]=x^2[/tex][tex]\begin{gathered} 121x^4y^6-25x^2z^{12} \\ x^2(\frac{121x^4y^6}{x^2}-\frac{25x^2z^{12}}{x^2}) \\ x^2(121x^2y^6-25z^{12}) \\ \text{note:} \\ \frac{x^4}{x^2}=x^{4-2}=x^2 \\ \frac{x^2}{x^2}=x^{2-2}=x^0=1 \end{gathered}[/tex]Step 2:
Expand the bracket using a figure of two squares
[tex]\begin{gathered} x^2(121x^2y^6-25z^{12}) \\ 121x^2y^6=(11xy^3)^2,25z^{12}=(5z^6)^2 \\ By\text{ substituting the expressions, we will have} \\ x^2(121x^2y^6-25z^{12})=x^2((11xy^3)^2-(5z^6)^2) \end{gathered}[/tex]Step 3:
Apply the difference of two squares principle below
[tex]a^2-b^2=(a-b)(a+b)[/tex][tex]x^2((11xy^3)^2-(5z^6)^2)=x^2(11xy^3-5z^6)(11xy^3+5z^6)[/tex]Hence,
The final answer is
[tex]x^2(11xy^3-5z^6)(11xy^3+5z^6)[/tex]