Question

Can you help me with questions no 13 only part b

Answer 1

To answer this question, we need to use the binomial theorem, and we have the identity when we raise the binomial to 4:

$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$

And we also have that:

$(a-b)^4=a^4-4a^3b_{}+6a^2b^2-4ab^3+b^4$

And now, we know that the values for a and b are:

$a=\sqrt[]{5},b=\sqrt[]{2}$

Then, we also know that we need the result for:

$(a+b)^4+(a-b)^4$

Then, if we substitute the equivalent expressions, we have:

$(a^4+4a^3b+6a^2b^2+4ab^3+b^4)+(a^4-4a^3b_{}+6a^2b^2-4ab^3+b^4)$

Simplifying, we have - we need to add the like terms as follows:

$a^4+a^4+4a^3b-4a^3b+6a^2b^2+6a^2b^2+4ab^3-4ab^3+b^4+b^4$

Therefore:

$2a^4+2(6a^2b^2)+2b^4=2a^4+12a^2b^2+2b^4$

We see that some terms were canceled since they are the same with opposite signs:

$\begin{gathered} 4a^3b-4a^3b=0 \\ 4ab^3-4ab^3=0 \end{gathered}$

Now, we can substitute the values for a and b as follows:

$a=\sqrt[]{5},b=\sqrt[]{2}$

We need to remember that:

$\sqrt[]{5}=5^{\frac{1}{2}},\sqrt[]{2}=2^{\frac{1}{2}}$

Then we have:

$2(5^{\frac{1}{2}})^4+12(5^{\frac{1}{2}})^2(2^{\frac{1}{2}})^2+2(2^{\frac{1}{2}})^4$

Thus

$2(5^{\frac{4}{2}})+12(5^{\frac{2}{2}})(2^{\frac{2}{2}})+2(2^{\frac{4}{2}})$$\begin{gathered} 2(5^2)+12(5)(2)+2(2^2) \\ 2(25)+12(10)+2(4) \\ 50+120+8=178 \end{gathered}$

In summary, therefore, we have that:

$(\sqrt[]{5}+\sqrt[]{2})^4+(\sqrt[]{5}-\sqrt[]{2})^4=178$