Expand x^n power of (x-1) and (y-1) up to third degree terms and hence evaluate (1.1) 1.02.

Steps to Solve

1. Expand the binomial (x-1) to the third degree

To expand the binomial expression $(x-1)^3$, we can use the binomial theorem, which states that:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

In our case, let $a = x$ and $b = -1$, and $n = 3$.

So, we write:

$$(x - 1)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} (-1)^k$$

Calculating each term gives:

For $k=0$: $\binom{3}{0} x^3 (-1)^0 = x^3$

For $k=1$: $\binom{3}{1} x^2 (-1)^1 = -3x^2$

For $k=2$: $\binom{3}{2} x^1 (-1)^2 = 3x$

For $k=3$: $\binom{3}{3} x^0 (-1)^3 = -1$

Combining these results, we have:

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

2. Expand the binomial (y-1) to the third degree

Similarly, we expand $(y-1)^3$ using the same approach:

$$(y - 1)^3 = \sum_{k=0}^{3} \binom{3}{k} y^{3-k} (-1)^k$$

Calculating the terms:

For $k=0$: $\binom{3}{0} y^3 (-1)^0 = y^3$

For $k=1$: $\binom{3}{1} y^2 (-1)^1 = -3y^2$

For $k=2$: $\binom{3}{2} y^1 (-1)^2 = 3y$

For $k=3$: $\binom{3}{3} y^0 (-1)^3 = -1$

Combining these results, we have:

$$(y - 1)^3 = y^3 - 3y^2 + 3y - 1$$

3. Evaluate the product of the two expansions at $(x=1.1, y=1.02)$

Now, we evaluate $(1.1 - 1)^3$ and $(1.02 - 1)^3$:

Let $x = 1.1$ and $y = 1.02$.

Calculate $(x - 1)^3 = (0.1)^3 = 0.001$

Calculate $(y - 1)^3 = (0.02)^3 = 0.000008$

Finally, we multiply these two results:

$$ (0.001) \cdot (0.000008) = 0.000000008 $$

4. Presenting the final result

Thus, the evaluated expression is:

$$ 0.000000008 $$