Question

What is a simple way to solve for the sum and difference of 2 cubes? For example, 27+64id="TexFormula1" title="x^3" alt="x^3" align="absmiddle" class="latex-formula"> and 64$m^3$-1

Answer 1

Answer:

(3 + 4x)(9 - 12x + 16x²) = 0

(4m - 1)(16m² + 4m + 1) = 0

Step-by-step 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 $- \frac{3}{4}$ 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 $\frac{1}{4}$ and the others we will get by applying Sridhar Acharya Formula, which will give a pair of conjugate imaginary roots of the equation.  (Answer)