Question

if a is equal to 1 by 3 minus 2 root 2 and b is equal to 3 + 2 root 2 prove that a square b + a b square is equal to 6​

Answer 1

Question :-

If $\bf{a = \dfrac{1}{3 - 2\sqrt{2}}}$ and

$\bf{b = \dfrac{1}{3 + 2\sqrt{2}}}$ , then prove that

$\bf{a^{2}b + ab^{2} = 6}$

To Find :-

To Prove that RHS = LHS.

Given :-

Value of a and b :

• $\bf{a = \dfrac{1}{3 - 2\sqrt{2}}}$

• $\bf{b = \dfrac{1}{3 + 2\sqrt{2}}}$

We Know :-

• $\bf{(a + b)^{2} = a^{2} + b^{2} + 2ab}$

• $\bf{(a - b)^{2} = a^{2} + b^{2} - 2ab}$

Solution :-

$\bf{a^{2}b + ab^{2} = 6}$

• LHS = a²b + ab²

• RHS = 6

First we have to solve the LHS of the Equation :-

By substituting the values of a and b in the Equation , we get :-

$:\implies \bf{a^{2}b + ab^{2}} \\ \\ \\ \\ :\implies \bf{\bigg(\dfrac{1}{3 - 2\sqrt{2}}\bigg)^{2} \times \dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{3 - 2\sqrt{2}} \times \bigg(\dfrac{1}{3 + 2\sqrt{2}}\bigg)^{2}}$

By Using the identity :-

• $\bf{(a + b)^{2} = a^{2} + b^{2} + 2ab}$
• $\bf{(a - b)^{2} = a^{2} + b^{2} - 2ab}$

We Get :-

$\implies \bf{\bigg(\dfrac{1}{3^{2} - 2 \times 3 \times 2\sqrt{2} + (2\sqrt{2})^{2}}\bigg) \times \dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{3 - 2\sqrt{2}} \times \bigg(\dfrac{1}{3^{2} + 2 \times 3 \times 2\sqrt{2} + (2\sqrt{2})}\bigg)^{2}}$

$:\implies \bf{\bigg(\dfrac{1}{9 - 6 \times 2\sqrt{2} + 4 \times 2}\bigg) \times \dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{3 - 2\sqrt{2}} \times \bigg(\dfrac{1}{9 + 6\times 2\sqrt{2} + 4 \times 2}\bigg)}$

$\implies \bf{\bigg(\dfrac{1}{9 - 12\sqrt{2} + 8}\bigg) \times \dfrac{1}{3 + 2\sqrt{2}} + \dfrac{1}{3 - 2\sqrt{2}} \times \bigg(\dfrac{1}{9 + 12\sqrt{2} + 8}\bigg)}$

$\implies \bf{\bigg(\dfrac{1}{(9 \times 3 - 12\sqrt{2} \times 3 + 8 \times 3) + (9 \times 2\sqrt{2} - 12\sqrt{2} \times 2\sqrt{2} + 8 \times 2\sqrt{2})}\bigg) + }$ $\bf{\bigg(\dfrac{1}{(9 \times 3 + 12\sqrt{2} \times 3 + 8 \times 3) + (9 \times - 2\sqrt{2} + 12\sqrt{2} \times - 2\sqrt{2} + 8 \times - 2\sqrt{2})}\bigg)}$

$\implies \bf{\bigg(\dfrac{1}{(27 - 35\sqrt{2} + 24) + (18\sqrt{2} - 48 + 16\sqrt{2})}\bigg) + \bigg(\dfrac{1}{(27 + 36\sqrt{2} + 24) + (-18\sqrt{2} - 48 - 16\sqrt{2})}\bigg)}$

$\implies \bf{\bigg(\dfrac{1}{(51 - 36\sqrt{2}) + (34\sqrt{2} - 48)}\bigg) + \bigg(\dfrac{1}{(51 + 36\sqrt{2}) + (-34\sqrt{2} - 48)}\bigg)}$

$\implies \bf{\bigg(\dfrac{1}{(51 - 36\sqrt{2}) + 34\sqrt{2} - 48)}\bigg) + \bigg(\dfrac{1}{(51 + 36\sqrt{2} -34\sqrt{2} - 48)}\bigg)}$

$\implies \bf{\bigg(\dfrac{1}{(3 - 2\sqrt{2})}\bigg) + \bigg(\dfrac{1}{(3 + 2\sqrt{2})}\bigg)}$

$:\implies \bf{\bigg(\dfrac{3 + 2\sqrt{2} + 3 - 2\sqrt{2}}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}\bigg)}$

Using the identity :-

$\bf{a^{2} - b^{2} = (a + b)(a - b)}$

we get :

$:\implies \bf{\bigg(\dfrac{6}{(3)^{2} - (2\sqrt{2})^{2}}\bigg)}$

$:\implies \bf{\bigg(\dfrac{6}{9 - 8}\bigg)}$

$:\implies \bf{\bigg(\dfrac{6}{1}\bigg)}$

$:\implies \bf{6}$

$\therefore \purple{\bf{LHS = 6}}$

Now putting the LHS and RHS together , we get :

$:\implies \purple{\bf{6 = 6}}$

Hence , RHS = LHS .

Proved :