Rationalising factor of cube root of a+cube root of b is...
Answers
Answer:
math]x^6-y^6=(x^3)^2-(y^3)^2[/math]
[math]=(x^3+y^3)(x^3-y^3)[/math]
[math]=(x+y)(x^2+y^2-xy)(x-y)(x^2+y^2+xy)[/math]
[math]=(x+y)(x-y)(x^2+y^2+xy)(x^2+y^2-xy)[/math]
[math]=(x+y)(x-y)((x^2+y^2)^2-(xy)^2)[/math]
[math]=(x+y)(x-y)(x^4+y^4+x^2y^2)[/math]
[math]=(x+y)(x-y)(x^4+y^4+x^2y^2)[/math]
Now, substituting [math]x=3^{1/2}[/math] & [math]y=5^{1/3}[/math]
[math](3^{1/2})^6-(5^{1/3})^6=(3^{1/2}+5^{1/3})(3^{1/2}-5^{1/3})((3^{1/2})^4+(5^{1/3})^4+(3^{1/2})^2(5^{1/3})^2)[/math]
[math]3^3-5^2=(3^{1/2}+5^{1/3})(3^{1/2}-5^{1/3})(9+5^{4/3}+(3)(5^{2/3}))[/math]
[math]2=(3^{1/2}+5^{1/3})(3^{1/2}-5^{1/3})(9+5^{4/3}+(3)(5^{2/3}))[/math]
[math]2=(3^{1/2}+5^{1/3})\underbrace{(3^{1/2}-5^{1/3})(9+5^{4/3}+3\cdot 5^{2/3})}[/math]
Hence, the rationalizing factor of [math](3^{1/2}+5^{1/3})[/math] is [math]\underline{(3^{1/2}-5^{1/3})(9+5^{4/3}+3\cdot 5^{2/3})}