Question

If x²-y²= 12 and x³+y³ = 72, find the value of x and y​

Answer 1

Given,

x³+y³ = 72,

From above equation,

y^3 = 72 - x^3

Also given,

x^2 - y^2 = 12

From above equation,

y^2 = x^2 - 12

We can write the above equation as,

x^2 - 12 = (72-x^3)^2/3

Or

(x^2-12)^3 = (72-x^3)^2

0 = x^4 - 4*x^3- 12*x^2 + 192.... (1)

The above equation is satisfied by 4,

As (x-4) is the factor of eq (1),

Then,

Solving above equation,

(x-4)(x^3 - 12x^2 + 192) = 0

Now either x - 4 = 0

Or x^3 - 12x^2 + 192 = 0

When x = 4

Then y from given equations y = 2.

Above equation has only one real solution so 2nd equation may be neglected.

Answer 2

Answer:

x= 4 & y = 2

Step-by-step explanation:

x²-y²= 12 and x³+y³ = 72

x² - y² = 12

=> (x +y)(x-y) = 12

either x+y & x-y both would be -ve  or

x+y & x -y both would be +ve

x + y can not be negative as x³ + y³ = 72 and positive

so it means x +y & x-y are positive

12 = 1 * 12

12 = 2* 6

12 = 3*4

case 1

x +y = 12

x - y = 1

x = 13/2  & y = 11/2

then x³ + y³ = 441 ≠ 72

case 2

x +y = 6

x - y = 2

x = 4  & y = 2

x³ + y³ = 4³ + 2³ = 72

So x= 4 & y = 2

Case 3

x + y =4

x - y = 3

x = 7/2  y = 1/2

x³ + y³ = 43  ≠ 72