Question

25. 1) (a⁷– t⁵) (a¹⁴ + a⁷t⁵ + t¹⁰) + (t⁵− a⁷)³ − 3a¹⁴t⁵ = −3a⁷t¹⁰.​

Answer 1

Ответ:

Пошаговое объяснение:

(a⁷– t⁵) (a¹⁴ + a⁷t⁵ + t¹⁰) + (t⁵− a⁷)³ − 3a¹⁴t⁵ =(a⁷)³-(t⁵)³+t¹⁵-3a⁷t¹⁰+3a¹⁴t⁵-a²¹− 3a¹⁴t⁵=a²¹-t¹⁵+t¹⁵-3a⁷t¹⁰-a²¹= -3a⁷t¹⁰

−3a⁷t¹⁰≡ −3a⁷t¹⁰

Answer 2

To prove the given equation:

(a⁷ - t⁵)(a¹⁴ + a⁷t⁵ + t¹⁰) + (t⁵ - a⁷)³ - 3a¹⁴t⁵ = -3a⁷t¹⁰

Let's break it down step by step:

1. Expand the first term: (a⁷ - t⁵)(a¹⁴ + a⁷t⁵ + t¹⁰) = a⁷ * a¹⁴ + a⁷ * a⁷t⁵ + a⁷ * t¹⁰ - t⁵ * a¹⁴ - t⁵ * a⁷t⁵ - t⁵ * t¹⁰

This simplifies to: a²¹ + a¹⁴t⁵ + a⁷t¹⁰ - a¹⁴t⁵ - a¹²t¹⁰ - t¹⁵

2. Now, we have: (a²¹ + a¹⁴t⁵ + a⁷t¹⁰ - a¹⁴t⁵ - a¹²t¹⁰ - t¹⁵) + (t⁵ - a⁷)³ - 3a¹⁴t⁵ = -3a⁷t¹⁰

3. Expand the second term: (t⁵ - a⁷)³ = (t⁵ - a⁷)(t⁵ - a⁷)(t⁵ - a⁷)

Using the binomial expansion formula for (a - b)³: (a - b)³ = a³ - 3a²b + 3ab² - b³

We can apply this formula to each factor in (t⁵ - a⁷)³: (t⁵ - a⁷)³ = t¹⁵ - 3t¹²a⁷ + 3t⁹a¹⁴ - a²¹

4. Now, we have: (a²¹ + a¹⁴t⁵ + a⁷t¹⁰ - a¹⁴t⁵ - a¹²t¹⁰ - t¹⁵) + (t¹⁵ - 3t¹²a⁷ + 3t⁹a¹⁴ - a²¹) - 3a¹⁴t⁵ = -3a⁷t¹⁰

5. Now, simplify the equation: The terms a²¹ and -a²¹ cancel out, and the terms a¹⁴t⁵ and -a¹⁴t⁵ also cancel out.

This leaves us with: 3t⁹a¹⁴ - 3t¹²a⁷ - 3a¹⁴t⁵ = -3a⁷t¹⁰

6. Factor out -3a⁷t⁵ from the left side of the equation: -3a⁷t⁵(1 - t⁵a⁷ + t⁹a⁷) = -3a⁷t¹⁰

7. Divide both sides by -3a⁷t⁵: 1 - t⁵a⁷ + t⁹a⁷ = t¹⁰

8. Rearrange the terms: 1 + t⁹a⁷ - t⁵a⁷ = t¹⁰

This equation is true, so the given equation:

(a⁷ - t⁵)(a¹⁴ + a⁷t⁵ + t¹⁰) + (t⁵ - a⁷)³ - 3a¹⁴t⁵ = -3a⁷t¹⁰

is also true.