Question

Solve the equation ,

$y^3 = x^3 + 8x^2 - 6x + 8$

for positive integers x and y .

​

Answer 1

Answer:

Original equation:

y3=x3+8∗x2−6∗x+8 -- (1)

Subtract equation-(1) by (x−1)3 on both sides,

We have,

y3−(x−1)3=(x3+8∗x2−6∗x+8)−(x−1)3

=5∗x2−9∗x+7 -- (2)

Consider the following quadratic equation (from RHS of equation-(2)) ,

5∗x2−9∗x+7=0

Its discriminant is 92−4∗5∗7=−59, which is less than zero. Now, from the theory of quadratic equations, we know that negative discriminant indicates that the above quadratic expression is positive for every real value of x . Hence,

5∗x2−9∗x+7>0 for every real x

From equation - (2),

=>y3−(x−1)3>0 for every real x

=>(x+1)<y -- (3)

Similarly, subtracting both sides of equation-(1) by (x+3)3 fetches the following:

(x+3)3−y3=(x+3)3−(x3+8∗x2−6∗x+8)

=x2+33∗x+19 -- (4)

x2+33∗x+19 has a positive derivative w.r.t x for x>0 i.e, (x+33). Hence, it is a monotonically increasing expression for x>0 . At x=0 , the value of x2+33∗x+19 is 19 ( >0 ). Hence, x2+33∗x+19>0 for x>0 and from equation-(4),

(x+3)3−y3>0

=>(x+3)>y -- (5)

From equations (3) and (5),

(x+1)<y<(x+3)

As we seek a positive integral solution,

y must be equal to (x+2).

Now substituting y=(x+2) in equation-(1), we get

2∗x2−18∗x=0

Two solutions:

• x=0 and y=2
• x=9 and y=11.

Answer 2

Solution:-Can be solved by simple.method

given that:-y^3=x^3+8x^2+8

here analysing we get

(x+2)^3+2x^2-18x=x^3+8+6x^2+12x+2x^2-18x=x^3+8x^2-6x+8

so,

y^3=x^3+8x^2+8=(x+2)^3+2x^2-18x

y^3+0=(x+2)^3+2x^2-18x

comparing both side we get

y^3=(x+2)^3......i)

and,0=2x^2-18x

2x^2=18x........ii)

x(2x-18)=0

x=0,x=9

now it said that,x is positive only,

so,ignore x=0

because 0 is neither positive nor negative

hence,x=9

putting this value in i) we get

y^3=(11)^3

y=11

hence,x=9 and y=11

{hope it helps}