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

 Jun 6, 2024
 #1
avatar+1952 
+1

We get two seperate equations from the problem. 

 

We have

a3+3ab2=6793a3ab2=615

 

Let's multiply the second equation by 3 so that we can get 3a^3 in both equations

3a3+9a2=2037

3a3ab2=615

 

Now, subtract the second equation from the first equation. We get

10ab2=1422ab2=142.2

 

Now, sub this value back into the first equation to get that

a3+3(142.2)=679a3+426.6=679252.4=a3a=3252.4a6.19

 

Now we have a, we can find b. 

b2=(142.2)/(6.19)b222.97b22.97b4.79

 

So, we have

 ab6.194.79ab1.40

 

So 1.4 is our answer

 

Thanks! :)

 Jun 6, 2024

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