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Help how do I solve this system

 

Let a and b be real numbers such that a^3 + 3ab^2 = 679 and a^3 - 3ab^2 = 673.  Find a - b.

 Feb 5, 2023
 #1
avatar+161 
0

Well lets do some algebra!

\(a^3 + 3ab^2 = 679\)

\(a^3 - 3ab^2 = 673\)

 

\(a^3 + 3ab^2 = 679 \) this is equal to

\(a^3 - 3ab^2 = 679 - 6ab^2\)

we can conclude that

\(673 = 679 - 6ab^2\)

\(6 = 6ab^2\)

\(ab^2 = 1\)

Most likely a will be very big because we need a^3 to equal something large since the second term is = to 3.

so a^3 = 676 so a = \(\sqrt[3]{676}\)

b = \(\sqrt{\frac{1}{\sqrt[3]{676}}},-\sqrt{\frac{1}{\sqrt[3]{676}}}\)

a - b = \(\sqrt[3]{676}-\frac{1}{\sqrt{\sqrt[3]{676}}},\sqrt[3]{676}+\frac{1}{\sqrt{\sqrt[3]{676}}}\) that is so definenly incorrect, so plz someone see if this is correct or not

 Feb 6, 2023
 #2
avatar+33574 
+1

1.  Add the two equations to get \(2a^3=1352\),  so  \(a=26^{2/3}\)

 

2. Subtract the two equations to get \(6ab^2=6\)  so \(ab^2=1\)

 

Hence \(b^2=26^{-2/3}\)  and \(b=\pm26^{-1/3}\)

 

Take it from there!

 Feb 6, 2023

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