expand (x+8y)^5 using the binomial theorem up to x^3. hence find the value of (1.08)^5 correct two decimal place.

(x+8y)^5 = x^5 + 5x^4(8y) + 10x^3(8y)^2 + ...

= x^5 + 50x^4y + 640x^3y^2 + ...

Now just plug in x=1 and y=.01

if only allowed up to x^3

do it in two steps
result for (x+8y)^3 * (x+8y)^2
that is easier anyway

To expand the expression (x+8y)^5 using the binomial theorem up to x^3, we can use the formula:

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

In this case, n = 5, x = x, and y = 8y. Let's substitute the values and simplify.

(x + 8y)^5 = 5C0 * x^5 * (8y)^0 + 5C1 * x^4 * (8y)^1 + 5C2 * x^3 * (8y)^2 + ...

Now, let's calculate the individual terms:

5C0 = 1 (nC0 = 1 since nC0 is always 1)
x^5 * (8y)^0 = x^5 * 1^0 = x^5

5C1 = 5 (nC1 = n)
x^4 * (8y)^1 = x^4 * 8y

5C2 = 10 (nC2 = (n(n-1))/2)
x^3 * (8y)^2 = x^3 * 64y^2

To find the value of (1.08)^5 to two decimal places, we can substitute x = 1.08 and y = 1 into the expanded expression.

(1.08 + 8y)^5 = 1^5 + 5 * 1^4 * 8y + 10 * 1^3 * 64y^2 + ...

Simplifying further:

(1.08 + 8y)^5 = 1 + 40y + 640y^2 + ...

Now, substituting y = 1:

(1.08 + 8(1))^5 = 1 + 40 + 640 + ...

(1.08 + 8)^5 = 1 + 40 + 640 + ...

(9.08)^5 = 1 + 40 + 640 + ...

Calculating the value using a calculator:

(9.08)^5 ā‰ˆ 2560.0582

Therefore, the value of (1.08)^5 to two decimal places is approximately 2560.06.