To prove that $E = a + a^2 + a^3 + b + b^2 + b^3$ under the given conditions:

Step 1: Simplify the given conditions

We are provided with: $a + b - 6 = ab - 1 = 1.$

From the first equality: $a + b - 6 = 1 \implies a + b = 7.$

From the second equality: $ab - 1 = 1 \implies ab = 2.$

Step 2: Use these results to calculate $a$ and $b$

The equations $a + b = 7$ and $ab = 2$ correspond to the roots of the quadratic equation: $x^2 - (a+b)x + ab = 0.$

Substituting $a+b = 7$ and $ab = 2$, the quadratic equation becomes: $x^2 - 7x + 2 = 0.$

Solving for $x$ using the quadratic formula: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(2)}}{2(1)} = \frac{7 \pm \sqrt{49 - 8}}{2} = \frac{7 \pm \sqrt{41}}{2}.$

Thus: $a = \frac{7 + \sqrt{41}}{2}, \quad b = \frac{7 - \sqrt{41}}{2}.$

Step 3: Compute $E = a + a^2 + a^3 + b + b^2 + b^3$

We separate $E$ into two components: $E = (a + b) + (a^2 + b^2) + (a^3 + b^3).$

Component 1: $a + b$

From earlier: $a + b = 7.$

Component 2: $a^2 + b^2$

Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$: $a^2 + b^2 = 7^2 - 2(2) = 49 - 4 = 45.$

Component 3: $a^3 + b^3$

Using the identity $a^3 + b^3 = (a+b)((a+b)^2 - 3ab)$: $a^3 + b^3 = 7(7^2 - 3(2)) = 7(49 - 6) = 7(43) = 301.$

Step 4: Combine the results

Adding all the components together: $E = (a + b) + (a^2 + b^2) + (a^3 + b^3),$ $E = 7 + 45 + 301 = 353.$

Conclusion

The value of $E$ is: $E = 353.$

Would you like detailed derivations of the intermediate steps or explanations for the formulas used?

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Related Questions:

1. What is the discriminant and how does it help solve quadratic equations?
2. How is the identity for $a^2 + b^2 = (a+b)^2 - 2ab$ derived?
3. What are other common ways to compute $a^3 + b^3$?
4. Can $a$ and $b$ be expressed explicitly in terms of radicals?
5. How can this problem be solved numerically or graphically?

Tip:

Always verify the roots of a quadratic equation by substituting back into the original equation to avoid errors.