If x,y > 0 and the x³y³ term of

(x + y)^6 is 20, then y^-3 equals:

To find the value of y^(-3), we need to first expand the expression (x + y)^6 using the binomial theorem. The general form of the binomial theorem is:

(x + y)^n = nC0 * x^n * y^0 + nC1 * x^(n-1) * y^1 + nC2 * x^(n-2) * y^2 + ... + nCr * x^(n-r) * y^r + ... + nCn * x^0 * y^n

Here, nC0, nC1, nC2, ..., nCr are the binomial coefficients, calculated as n!/((n-r)! * r!).

In the given expression, (x + y)^6, we need to find the term with x^3 * y^3, which corresponds to r = 3. The term will be:

6C3 * x^(6-3) * y^3 = 6C3 * x^3 * y^3

According to the problem, this term is equal to 20:

6C3 * x^3 * y^3 = 20

Now, let's solve this equation to find the value of y:

6C3 = 6!/(3!(6-3)!) = 6!/(3! * 3!) = 20 (notice that 6C3 = 6C3 = 20)

20 * x^3 * y^3 = 20

Simplifying the equation, we find:

x^3 * y^3 = 1

To find the value of y^(-3), we can rewrite y^3 as (y^(-1))^(-3):

x^3 * (y^(-1))^(-3) = 1

Since we know that x, y > 0, we can rewrite the equation as:

1/(x^3 * y)^(3) = 1

Taking the cube root of both sides, we have:

1/(x^3 * y) = 1

Now we can solve this equation for y:

x^3 * y = 1

y = 1/(x^3)

Therefore, y^(-3) is equal to (1/(x^3))^(-3):

y^(-3) = (1/(x^3))^(-3)
= (x^3)^3
= x^9

So, y^(-3) is equal to x^9.