Question

x4 + x2y2 + y4 = 21
x2 + xy + y2 = 7
1/x^2 + 1/y^2= ?

Answer 1

Answer:

I think it is in the form of x+y square form so answer is 1 by 7

Answer 2

Answer:

Step-by-step explanation:

x⁴+x²y²+y⁴ = 21

Now, add and subtract on LHS by x²y²

(x⁴+x²y²+y⁴)+x²y²-x²y²= 21

(x⁴+2x²y²+y⁴) - x²y² =21

(x²+y²)² - (xy)² = 21

(x²+y²+xy)(x²+y²-xy) = 21

But, (x²+y²+xy) = 7 ----------------> (i)

So, (x²+y²-xy) = 21/7 = 3 ----------------> (ii)

Now,

we can also write,

1/x² + 1/y² = (x²+y²)/x²y²

To find (x²+y²),

We can add eq (i) & (ii)

(x²+y²+xy) + (x²+y²-xy) = 7+3 = 10

2(x²+y²) = 10

(x²+y²) = 10/2 = 5

Now find x²y²,

We can subtract eq (i) - (ii),

(x²+y²+xy) - (x²+y²-xy) = 7 - 3 = 4

2(xy) = 4

xy = 2

Therefore, x²y² = 2² = 4

So final answer is, (x²+y²)/x²y² = 5/4