We are given the following system of equations for two numbers $a$ and $b$:

$a + b = 2$ $a^2 + b^2 = 6$

Part (a): Find the value of $a^3 + b^3$.

We can use the identity for the sum of cubes:

$a^3 + b^3 = (a + b)\left(a^2 - ab + b^2\right)$

From the first equation, we know:

$a + b = 2$

Next, to find $ab$, we use the identity:

$a^2 + b^2 = (a + b)^2 - 2ab$

Substituting $a + b = 2$ and $a^2 + b^2 = 6$:

$6 = 2^2 - 2ab$

$6 = 4 - 2ab$

$2ab = -2$

$ab = -1$

Now, substituting $a + b = 2$, $ab = -1$, and $a^2 + b^2 = 6$ into the expression for $a^3 + b^3$:

$a^3 + b^3 = (a + b)\left(a^2 - ab + b^2\right) = 2 \times (6 - (-1)) = 2 \times 7 = 14$

Thus, the value of $a^3 + b^3$ is:

$a^3 + b^3 = 14$

Part (b): Find the value of $a^5 + b^5$.

We will use the identity:

$a^5 + b^5 = (a + b)(a^4 - ab(a^3 + b^3) + b^4)$

We need to compute $a^4 + b^4$. Using the identity for the fourth powers:

$a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2$

We already know that $a^2 + b^2 = 6$ and $ab = -1$, so:

$a^4 + b^4 = 6^2 - 2(-1)^2 = 36 - 2 = 34$

Now, using the expression for $a^5 + b^5$:

$a^5 + b^5 = (a + b)\left(a^4 - ab(a^3 + b^3) + b^4\right)$

Substituting $a + b = 2$, $ab = -1$, $a^3 + b^3 = 14$, and $a^4 + b^4 = 34$:

$a^5 + b^5 = 2 \times (34 - (-1)(14)) = 2 \times (34 + 14) = 2 \times 48 = 96$

Thus, the value of $a^5 + b^5$ is:

$a^5 + b^5 = 96$

Summary:

• $a^3 + b^3 = 14$
• $a^5 + b^5 = 96$

Would you like more details on any step?

Here are 5 related questions:

1. How would you solve for $a$ and $b$ individually in this system?
2. What is the general formula for $a^n + b^n$ in terms of lower powers of $a$ and $b$?
3. Can you express $a^7 + b^7$ using a similar approach?
4. How can these methods be extended to complex numbers?
5. How does the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ generalize?

Tip: Using algebraic identities simplifies the computation of higher powers, reducing the need to solve for $a$ and $b$ explicitly.