If abc 0, what is the value of a*b*c ?

Given:
\((a + b)^{10} = c\)
\((a + c)^{10} = b\)
\((b + c)^{10} = a\)

Asked: If \(abc \neq 0\), what is the value of a*b*c ?

Since the equations are similar w.r.t a, b & c.
We can conclude a = b = c

\((a + a)^{10} = a\)
\(2^10*a^10 = a\)
\(a^9 = \frac{1}{2^10[}/m]\\
[m]a = \frac{1}{2^{10/9}}\)
\(a*b*c = a^3 = \frac{1}{2^{10/9*3} }= \frac{1}{2^{10/3}} = \frac{1}{2^32^{1/3}} = \frac{1}{8*\sqrt[3]{2}}}\)

IMO B

Official Solution:


\((a + b)^{10} = c\)

\((a + c)^{10} = b\)

\((b + c)^{10} = a\)

If \(abc \neq 0\), what is the value of \(a*b*c\)?


A. \(\frac{1}{8*\sqrt[10]{8}}\)
B. \(\frac{1}{8*\sqrt[3]{2}}\)
C. \(\frac{1}{6*\sqrt[3]{2}}\)
D. \(\frac{1}{4*\sqrt[3]{2}}\)
E. \(\frac{1}{2*\sqrt[3]{2}}\)


First of all, notice that since \(a\), \(b\), and \(c\), equal to an even power of nonzero numbers, then all of them must be positive.

Next, realize that it must be true that \(a=b=c\). Why? Let's assume that \(b > c\). Now, if \(b > c\) (and since \(a > 0\), \(b > 0\) and \(c > 0\)), then \((a + b)^{10} > (a + c)^{10} \). But it's given that \((a + b)^{10} = c\) and \((a + c)^{10} = b\), so it would mean that \(c > b\). This contradicts the initial assumption that \(b > c\), so it cannot be true. Similarly you can prove that \(b < c\) cannot be true, which would mean that \(b = c\). Using the same logic you can further get that \(a = b\), and therefore \(a=b=c\).

Since \(a=b=c\), our question becomes: what is the value of \(a*b*c=a*a*a=a^3\)?

Substitute \(b\) and \(c\) in any equation given by \(a\) to get \((a + a)^{10} = a\):

\((2a )^{10} = a\)

\(2^{10}*a^{10} = a\) (distributing the power);

\(2^{10}*a^{9} = 1\) (dividing by nonzero \(a\));

\(a^9 = \frac{1}{2^{10}}\)

Since we need the value of \(a^3\), take the cube root of the above:

\(a^3 = \frac{1}{\sqrt[3]{2^{10}}}\);

\(a^3 = \frac{1}{\sqrt[3]{2^9*2}}\);

\(a^3 = \frac{1}{2^3\sqrt[3]{2}}\);

\(a^3 = \frac{1}{8\sqrt[3]{2}}\).

Answer: B

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