Question

(х+2)в в кубе(3-х)в кубе(2а -в)в кубе​

Answer 1

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Answer 2

To simplify the expression, let's expand each cube term:

1. Expand (х + 2)³: (х + 2)³ = (х + 2)(х + 2)(х + 2) = (х + 2)(х² + 2х + 2х + 4) = (х + 2)(х² + 4х + 4) = х³ + 4х² + 4х + 2х² + 8х + 8 = х³ + 6х² + 12х + 8

2. Expand (3 - х)³: (3 - х)³ = (3 - х)(3 - х)(3 - х) = (3 - х)(9 - 6х + х²) = 27 - 18х + 3х² - 9х + 6х² - х³ = 27 - 27х + 9х² - х³

3. Expand (2а - в)³: (2а - в)³ = (2а - в)(2а - в)(2а - в) = (2а - в)(4а² - 4ав + в²) = 8а³ - 8а²в - 8а²в + 8ав² + 2ав² - 2в³ = 8а³ - 16а²в + 10ав² - 2в³

Now, we multiply all the expanded terms together:

(х + 2)³(3 - х)³(2а - в)³ = (х³ + 6х² + 12х + 8)(27 - 27х + 9х² - х³)(8а³ - 16а²в + 10ав² - 2в³)

To simplify further, we can multiply these terms out by expanding the brackets and combining like terms. However, the result would be quite lengthy and might not have a specific factorization. If you have a specific value for 'х' and 'a,' I can help you calculate the final value. Otherwise, I hope this intermediate result helps you understand the expression's expansion.