Question

2a6-128a3/(2a2+8a+32)*(a4-4a3) упростить пж

Answer 1

$\frac{2 {a}^{6} - 128 {a}^{3} }{(2 {a}^{2} + 8 a + 32)( {a}^{4} - 4 {a}^{3}) } = \frac{ 2{a}^{3}({a}^{3} - 64 {a}^{3} ) }{2( {a}^{2} + 4a + 16) \times {a}^{3} (a - 4)} = \frac{2(a - 4)( {a}^{2} + 4a + 16)}{2( {a}^{2} + 4a + 16)(a - 4)} = 1$

Answer 2

To simplify the expression 2a^6 - 128a^3 / (2a^2 + 8a + 32) * (a^4 - 4a^3), we can follow the order of operations (PEMDAS/BODMAS) and simplify each part of the expression step by step.

Let's break it down:

Step 1: Simplify the numerator

The numerator is 2a^6 - 128a^3. We can't simplify it further because there are no like terms to combine.

Step 2: Simplify the denominator

The denominator is (2a^2 + 8a + 32). Again, we can't simplify it further because there are no like terms to combine.

Step 3: Simplify the expression inside the parentheses

The expression inside the parentheses is (a^4 - 4a^3). We can't simplify it further because there are no like terms to combine.

Step 4: Simplify the division

To simplify the division, we need to multiply the numerator by the reciprocal of the denominator. The reciprocal of (2a^2 + 8a + 32) is 1 / (2a^2 + 8a + 32).

So, the expression becomes:

(2a^6 - 128a^3) * (1 / (2a^2 + 8a + 32)) * (a^4 - 4a^3)

Step 5: Simplify the multiplication

To simplify the multiplication, we can cancel out common factors between the numerator and the denominator.

Let's factor out common terms from the numerator and the denominator:

Numerator: 2a^6 - 128a^3 = 2a^3(a^3 - 64) = 2a^3(a - 4)(a^2 + 4a + 16)

Denominator: 2a^2 + 8a + 32 = 2(a^2 + 4a + 16)

Now, we can cancel out the common factor of 2(a^2 + 4a + 16):

(2a^3(a - 4)(a^2 + 4a + 16)) / (2(a^2 + 4a + 16)) * (a^4 - 4a^3)

Step 6: Simplify the expression

After canceling out the common factor, we are left with:

a^3(a - 4) * (a^4 - 4a^3)

Now, we can simplify further by multiplying the terms:

a^3 * a^4 = a^7

a^3 * (-4a^3) = -4a^6

So, the final simplified expression is:

a^7 - 4a^6(a - 4)

I hope this helps! Let me know if you have any further questions.