Question

При a-b=2, a3-b3=28, решить a2+ab+b2​

Answer 1

$a-b=2\\a^{3}-b^{3}=28\\a^{2}+ab+b^{2}=?\\\\a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})\\28=2(a^{2}+ab+b^{2})\\a^{2}+ab+b^{2}=\frac{28}{2}=14$

Answer 2

To solve the equation, let's express a^3 - b^3 as (a - b)(a^2 + ab + b^2). We already know that a - b = 2, so we can substitute this value:

(a^3 - b^3) = (a - b)(a^2 + ab + b^2) 28 = 2(a^2 + ab + b^2)

Dividing both sides of the equation by 2, we get:

14 = a^2 + ab + b^2

Now let's square the equation a - b = 2 to find an expression for a^2 + 2ab + b^2:

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

Next, let's add this equation to the previous equation:

(a^2 + ab + b^2) + (a^2 - 2ab + b^2) = 14 + 4 2a^2 + 2b^2 = 18

Simplifying the equation further:

a^2 + b^2 = 9

Now we have a system of two equations:

1. a^2 + ab + b^2 = 14
2. a^2 + b^2 = 9

Subtracting equation 2 from equation 1, we can eliminate the a^2 term:

(a^2 + ab + b^2) - (a^2 + b^2) = 14 - 9 ab = 5

Now we have the value of ab, and we can substitute it back into equation 2:

a^2 + b^2 = 9 a^2 + (ab)^2 = 9 a^2 + 25 = 9 a^2 = 9 - 25 a^2 = -16

Since a^2 is negative, there are no real solutions for a. Therefore, there is no real value for a^2 + ab + b^2.