Question

27n⁶-64=(3n²)³-4³= 9x²-1=(3x)²-1²

Answer 1

27n⁶-64=(3n²)³-4³=(3n²-4)(9n⁴+12n²+16)

9x²-1=(3x)²-1²=(3x-1)(3x+1)

Answer 2

It appears that you've provided two different equations:

1. $27n^6 - 64 = (3n^2)^3 - 4^3$
2. $9x^2 - 1 = (3x)^2 - 1^2$

Let's solve each equation separately:

For Equation 1:

$27n^6 - 64 = (3n^2)^3 - 4^3$

First, simplify the right side:

$3n^2 = 3^1 * n^2$

$(3n^2)^3 = (3^1 * n^2)^3 = 3^3 * n^6 = 27n^6$

So the equation becomes:

$27n^6 - 64 = 27n^6 - 64$

Now, you can see that the equation is already simplified, and there is no need to solve it further. The solution to this equation is any value of $n$ that satisfies it.

For Equation 2:

$9x^2 - 1 = (3x)^2 - 1^2$

Now, simplify both sides of the equation:

$(3x)^2 = 3^2 * x^2 = 9x^2$

So the equation becomes:

$9x^2 - 1 = 9x^2 - 1$

Again, you can see that the equation is already simplified, and the solution to this equation is any value of $x$ that satisfies it.

In both cases, the equations are identities, which means they are true for all values of $n$ in the first equation and for all values of $x$ in the second equation.