### Question 22

SSC-CGL 2020 Mar 3 Shift 1

If x⁴ + x²y² + y⁴ = 21, x² + xy + y² = 7, then the value of (1/x² + 1/y²) = ?

### Solution in Brief

Factorize x⁴ + x²y² + y⁴ = (x² + xy + y²)(x² - xy + y²), to get x² - xy + y² = 3. Subtract it from x² + xy + y² = 7 to get xy = 2. Make a guess x = 1, y = 2 to get 5/4 as the answer!

### Solution in Detail

Given $\displaystyle x^2 + xy + y^2 = 7$

$\displaystyle \therefore (x^2 + y^2) = 7 - xy\text{ . . . (1)}$

Also given $\displaystyle x^4 + x^2y^2 + y^4 = 21$

Add $\displaystyle x^2y^2$ to both sides:

$\displaystyle x^4 + 2x^2y^2 + y^4 = 21 + x^2y^2$

Completing the squares

$\displaystyle (x^2 + y^2)^2 = 21 + x^2y^2$

Putting $\displaystyle x^2 + y^2$ from (1),

$\displaystyle (7 - xy)^2 = 21 + x^2y^2$

$\displaystyle \implies xy = 2$

Put in (1), $\displaystyle x^2 + y^2 = 7 - 2 = 5$

To find $\displaystyle \frac{1}{x^2} + \frac{1}{y^2}$

$\displaystyle = \frac{x^2 + y^2}{(xy)^2} = \frac{5}{2^2} = \frac54\:\underline{Ans}$

### More Solved Papers

This Blog Post/Article "(solved)Question 22 SSC-CGL 2020 March 3 Shift 1" by Parveen is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.