Find the positive integers c and d such that the cube root of 99-(70sqrt 2) = c - d sqrt 2?

Question

Please advise, thanks.

Hi,
Can someone help out and explain please.. thanks.

Answer 1

If you cube both sides you get
99 - 70sqrt2 = (c-dsqrt2)^3
so 99 - 70sqrt2 = (c^3+6c(d^2)) - sqrt2 * (3(c^2)d+2(d^3))
after grouping by terms with sqrt2 and without.
Since sqrt2 is not a real number, its integral multiples are also not real, so you can say from above that
99 = c^3+6c(d^2)
and
70 = 3(c^2)d+2(d^3)

Since c is a factor on top and d on the bottom, you can tell that c has to be an odd number (and multiple of 3) and that d then has to be even. It is easy then to see that c=3 and d = 2.

Answer 2

Hi,

Find the positive integers c and d such that the cube root of 99-(70sqrt 2) = c - d sqrt 2?

The problem statement is ambiguous
So I am solving for2 cases, meaning for two different problem statements.

Case 1:
99^1/3- 70√2 = c - d√2
as these involves surds, on equating coeffs, of √2
d = 70 and c = 99^1/3, 99^1/3 is not an integer

case 2:
(99 - 70√2)^1/3 = c - d√2

Raise the powers by 3 on both sides, to get
99 - 70√2 = c^3 - 2d^3√2 - 3c^2*d√2 + 6cd^2 ...(1)

on equating the coefficients of surd √2, we get
70 = 2d^3 + 3d*c^2 ............(2)
As c & d or integers, On cursory inspection, we get
c = 3 and d = 2 ................(3)

From (1), (2) and (3)
99 - 70√2 = 3^3 - 70√2 + 6*3*2^2
99 = 27 + 72

so, c = 3 and d = 2, which satify the given equation

Good luck

Answer 3

4 and 9

Answer 4

what?

i'm good at math...

but that...

that doesn't make any sense to me...