Question

Prove that
$(x^3-\frac{1}{x^3})(x^3+\frac{1}{x^3})(x^6+\frac{1}{x^6})=(x^{12}-\frac{1}{x^{12}})$

Answer 1

$(x^3-\frac{1}{x^3})(x^3+\frac{1}{x^3})(x^6+\frac{1}{x^6})$

we use

$(a-b)(a+b)=a^2-b^2$

here we can apply this for $1^{st}$ two temps

and we can write these two terms as

$(x^3-\frac{1}{x^3})(x^3+\frac{1}{x^3})=(x^3{)}^2-(\frac{1}{x^3}{)}^2$

$=x^6-\frac{1}{x^6}$

we put the value of $(x^3-\frac{1}{x^3})(x^3+\frac{1}{x^3})$ in

above eq.

$(x^6-\frac{1}{x^6})(x^6+\frac{1}{x^6})$

again we can apply above therm

Now we get $(x^6+\frac{1}{x^6})(x^6-\frac{1}{x^6})={\left(x^6\right)}^2-{\left(\frac{1}{x^6}\right)}^2$

$=x^{12}-\frac{1}{x^{12}}$

so $(x^3-\frac{1}{x^3})(x^3+\frac{1}{x^3})(x^6+\frac{1}{x^6})=x^{12}-\frac{1}{x^{12}}$