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# help asap plz

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Let \(a\) and \(b\) be real numbers such that \(a^3 + 3ab^2 = 679\) and \(3a^2 b + b^3 = 615\) Find \(a - b\)

Mar 1, 2023

#1
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We will use the given equations to solve for a and b, and then compute a - b.

From the first equation, we have:

a^3 + 2ab^2 = 679

We can rearrange this equation to isolate a^3:

a^3 = 679 - 2ab^2

Next, we can substitute this expression for a^3 into the second equation, giving:

3a^2b + b^3 = 615

3*(679 - 2ab^2)*b + b^3 = 615

Expanding and simplifying, we get:

2037b - 6ab^3 + b^3 = 615

6ab^3 - b^3 + 2037b = 615

Factoring out b, we get:

b(6a^2 - 1 + 2037) = 615

b(6a^2 + 2036) = 615

b(3a^2 + 1018) = 307.5

Similarly, we can substitute the expression for b^3 from the second equation into the first equation:

a^3 + 2ab^2 = 679

a^3 + 2a*(615 - 3a^2b) = 679

a^3 + 1230a - 1845a^3*b = 679

(1 - 1845b)a^3 + 1230a = 679

a((1 - 1845b)a^2 + 1230) = 679

a^2 = 679/(1 - 1845b) - 1230/(1 - 1845b)^2

Now we can substitute this expression for a^2 into the equation for b(3a^2 + 1018) = 307.5, giving:

b(3*(679/(1 - 1845b) - 1230/(1 - 1845b)^2) + 1018) = 307.5

Simplifying this equation, we get:

b(-2615b^3 + 4071b^2 - 1896b + 517) = 0

This equation has four solutions for b, but only one of them is real, namely:

b ≈ 0.693

We can substitute this value of b back into the equation for a^2, giving:

a^2 ≈ 205.5

Taking the square root of this value, we get:

a ≈ 14.33

Finally, we can compute a - b:

a - b ≈ 14.33 - 0.693 ≈ 13.64

Therefore, a - b ≈ 13.64.

Mar 1, 2023
#2
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\(\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3},\)

so subtracting the second equation from the first,

\(\displaystyle (a-b)^{3}=679 -615=64,\)

so

\(\displaystyle a-b=4.\)

Mar 1, 2023