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# Algebra

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157
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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
+266
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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
+33616
+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