What is the third term of (x^2 + 3y)^3
look at (a + b)^3 ... a^3 + 3 a^2 b + 3 a b^2 + b^3
just substitute
To find the third term of (x^2 + 3y)^3, we need to expand the expression using the binomial theorem. The binomial theorem states that for any two terms a and b raised to the power of n, the binomial expansion is given by:
(a + b)^n = nC0 * a^n * b^0 + nC1 * a^(n-1) * b^1 + nC2 * a^(n-2) * b^2 + ... + nCn-1 * a^1 * b^(n-1) + nCn * a^0 * b^n
In this case, a = x^2 and b = 3y, and we want to find the third term, which corresponds to the term with n = 3 and r = 2 (since the terms are indexed from zero).
To determine the coefficient, we can use the formula for combinations:
nCr = n! / (r! * (n-r)!)
where n! means the factorial of n.
Let's calculate each component:
nC2 = 3! / (2! * (3-2)!) = 3
Now, we can substitute these values into the binomial expansion formula to find the third term:
(x^2 + 3y)^3 = 3C2 * (x^2)^(3-2) * (3y)^2
Simplifying further, we have:
3 * x^(2*1) * 9y^2
= 27x^2y^2
Therefore, the third term of (x^2 + 3y)^3 is 27x^2y^2.