Question

What is a simple way to solve for the sum and difference of 2 cubes? For example, 27+64
`x^3` and 64
`m^3`-1

Answer 1

(3 + 4x)(9 - 12x + 16x²) = 0

(4m - 1)(16m² + 4m + 1) = 0

Explanation:

Here we have to solve the sum of two cubes which is
`27 + 64x^(3)`

Now, the equation is
`27 + 64x^(3) = 0`

⇒ 3³ + (4x)³ = 0

⇒ (3 + 4x)[3² - 3(4x) + (4x)²] = 0

⇒ (3 + 4x)(9 - 12x + 16x²) = 0

So, (3 + 4x) = 0 or (9 - 12x + 16x²) = 0

Therefore, from the above two relation we can solve for x.

One root will be
`- (3)/(4)` and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation.

Again, we have to solve the difference of two cubes which is
`64m^(3) - 1`

Now, the equation is
`64m^(3) - 1 = 0`

⇒ (4m)³ - 1³ = 0

⇒ (4m - 1)[(4m)² + 4m(1) + 1²] = 0

⇒ (4m - 1)(16m² + 4m + 1) = 0

So, (4m - 1) = 0 or (16m² + 4m + 1) = 0

Therefore, from the above two relation we can solve for m.

One root will be
`(1)/(4)` and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation. (Answer)