If a, b and x are integers such

Official Solution

• \(a^6=b^3=\frac{|x|}{x}\)

• If \(x > o, |x| = x, \frac{|x|}{x}=\frac{x}{x}=1\)

• If \(x < o, |x| = -x, \frac{|x|}{x}=−\frac{x}{x}=−1\)

• \(a^6=\frac{|x|}{x}\)

• As \(a^6\) is always positive,\(a^6 = 1\), i.e. \(a = 1\) or \(-1\)

• So, we can reject the value of \(\frac{|x|}{x}=−1\)

• \(b^3=\frac{|x|}{x}=1\)

• b = 1

• If \(a = 1\) and \(b = 1, a – b = 0\)

• If \(a = -1\) and \(b = 1, a- b = -2\)

• Rewriting \(a^3b^7\) as \(ab(a^2b^6)\)

• Therefore, \(ab(a^2b^6)>0\)

• We know that \(a^2b^6\) is always \(> 0\) (even power of any number is always positive)


• So, for \(ab(a^2b^6)> 0\)

• \(ab > 0\)

• This tells us that a and b have same signs.

• Since \(b > 0\), therefore a will also be greater than 0, so the value of \(a = 1\).


• \(a – b = 1 -1 = 0\)

• If \(a = 1\) and \(b = 1\), \(a + b = 2 > 0\)

• If \(a = -1\) and \(b = 1\), \(a + b = 0\), is not greater than zero


• Hence, we have a unique answer, where \(a =1\) and \(b = 1\)

• Thus \(a – b = 1 – 1 = 0\).

• If \(a = 1\) and \(b = 1, a – b = 0\)

• If \(a = -1\) and \(b = 1, a- b = -2\)