Question

if a = 1 by 3 minus 2 root 2 and b is equal to 1 by 3 + 2 root 2 find the value of a square b + a b square
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Answer 1

The value of $a^2b+ab^2=-(b+a)$

Step-by-step explanation:

Given expressions are $a=\frac{1}{3-2\sqrt{2}}$ and $b=\frac{1}{3+2\sqrt{2}}$

To find $a^2b+ab^2$

Substitute the values of a and b $a=\frac{1}{3-2\sqrt{2}}$ and $b=\frac{1}{3+2\sqrt{2}}$ in the above expression

$a^2b+ab^2=\left[\frac{1}{3-2\sqrt{2}}\right]^2\left[\frac{1}{3+2\sqrt{2}}\right]+\left[\frac{1}{3-2\sqrt{2}}\right]\left[\frac{1}{3+2\sqrt{2}}\right]^2$  ( using the identities $(a+b)^2-a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$ )

$=(\frac{1}{3^2-2(3)(2\sqrt{2})+(2\sqrt{2})^2)})\left[\frac{1}{3+2\sqrt{2}}\right]+\left[\frac{1}{3-2\sqrt{2}}\right](\frac{1}{3^2+2(3)(2\sqrt{2})+(2\sqrt{2})^2})$

$=\frac{1}{9-12\sqrt{2}+8})(\frac{1}{3+2\sqrt{2}})+(\frac{1}{3-2\sqrt{2}})(\frac{1}{9+12\sqrt{2}+8})$

$=\frac{1}{(9-12\sqrt{2}+8)(3+2\sqrt{2})}+\frac{1}{(3-2\sqrt{2})(9+12\sqrt{2}+8)}$ ( using the distributive property )

$=\frac{1}{27+18\sqrt{2}-36\sqrt{2}-48+24+16\sqrt{2}}+\frac{1}{27+36\sqrt{2}+24-18\sqrt{2}-48-16\sqrt{2}}$

$=\frac{1}{-3-2\sqrt{2}}+\frac{1}{-3+2\sqrt{2}}$

$=-\frac{1}{3+2\sqrt{2}}-\frac{1}{3-2\sqrt{2}}$

$=-b-a$

$=-(b+a)$

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

Answer 2

Answer:

$2 { \sqrt[..3y \\ 333 \times \frac{?}{?} ]{?} \times \frac{?}{?} }^{2}$