Algebra

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

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!