• Aug 29th 2011, 07:17 AM
grandepunto
• Aug 29th 2011, 07:25 AM
Siron
What have you tried?

1)
Use basic rules like: $\sqrt[n]{x^{a}}=x^{\frac{a}{n}}$ and $a^{x}\cdot a^{y}=a^{x+y}$
2)
The same over here, you have to know some basic rules like: $(a-b)\cdot (a+b)=a^2-b^2$ and $(a+b)^2=a^2+2ab+b^2$

Try to use them where necessary and simplify.
• Aug 29th 2011, 07:33 AM
grandepunto
thank you. i will give it another try.
• Aug 29th 2011, 07:40 AM
Siron
Ok! You can show your work afterwards if you want :).
• Aug 29th 2011, 07:54 AM
grandepunto
$\sqrt[4]{a^{3}}+\sqrt[4]{a}\cdot \sqrt{a}+\frac{3\sqrt[4]{a^{9}}}{\sqrt{a^{3}}}$
$=a^{\frac{3}{4}}+a^{\frac{1}{4}}\cdot a^{\frac{1}{2}}+\frac{3a^{\frac{9}{4}}}{a^{\frac{3 }{2}}}$
$=a^{\frac{3}{4}}+a^{\frac{3}{4}}+3a^{\frac{3}{4}}= 5a^{\frac{3}{4}}}=5\sqrt[3]{a^{4}}$
$(2\sqrt{7}-3)\cdot (2\sqrt{7}+3)-(\sqrt{7}+1)^2-(\sqrt{7}-2)^2$
$[(2\sqrt{7})^2-9]-(7+2\sqrt{7}+1}-(7-4\sqrt{7}+4)=28-9-7-2\sqrt{7}-1-7+4\sqrt{7}-4=2\sqrt{7}$