Question

ATTENTION••••
DO ONLY 5TH AND 6TH ..
I NEED YOUR HELP ...

Answer 1

We will use the binomial expansion theorem:

$(a + b)^{n} = \sum_{r=0}^{n} {n\choose r}a^{n-r}b^{r}.$

Now, for (v), we have:

$\left(\frac{1}{2}a^{2} + \frac{2}{3}b^{2}\right)^{3} = {3\choose 0}(\frac{1}{2}a^{2})^{3} +  {3\choose 1}(\frac{1}{2}a^{2})^{2} (\frac{2}{3}b^{2}) + {3\choose 2}(\frac{1}{2}a^{2})(\frac{2}{3}b^{2})^{2} + {3\choose 3}(\frac{2}{3}b^{2})^{3}$

$= \frac{1}{8}a^{6} + 3(\frac{1}{4}a^{4})(\frac{2}{3}b^{2}) + 3(\frac{1}{2}a^{2})(\frac{4}{9}b^{4}) + \frac{8}{27}b^{6}$

$= \underline{\underline{\frac{1}{8}a^{6} + \frac{1}{2}a^{4}b^{2} + \frac{2}{3}a^{2}b^{4} + \frac{8}{27}b^{6}}}.$

For (vi) we similarly have:

$\left(\frac{1}{3}a^{2}x^{3} - 2b^{3}y^{2}\right)^{3} = {3\choose 0}(\frac{1}{3}a^{2}x^{3})^{3} +  {3\choose 1}(\frac{1}{3}a^{2}x^{3})^{2} (-2b^{3}y^{2}) + {3\choose 2}(\frac{1}{3}a^{2}x^{3})(-2b^{3}y^{2})^{2} + {3\choose 3}(-2b^{3}y^{2})^{3}$

$= \frac{1}{27}a^{6}x^{9} + 3(\frac{1}{9}a^{4}x^{6})(-2b^{3}y^{2}) + 3(\frac{1}{3}a^{2}x^{3})(4b^{6}y^{4}) - 8b^{9}y^{6}$

$= \underline{\underline{\frac{1}{27}a^{6}x^{9} - \frac{2}{3}a^{4}b^{3}x^{6}y^{2} + 4a^{2}b^{6}x^{3}y^{4} - 8b^{9}y^{6}}}.$