Question

The difference between two positive integers is 2 and the difference between their cubes is 56. Find the number.

Answer 1

Hint: Suppose two positive integers in terms of a and b, then according to the question statement form two equations and solve to get an answer.

Complete step-by-step answer:
Let the two numbers be a and b
According to the question
a – b = 2 ––––––––I
Or a = b + 2
And \[{a^3}-{b^3} = {\text{ }}56\]––––––––II
Put the value of a in equation II
\[{\left( {b + 2} \right)^3}-{b^3} = {\text{ }}56\]
Applying \[{\left( {a{\text{ }} +{\text{ }}b} \right)^3}\] formula i.e \[\left( {a{\text{ }} + {\text{ }}b} \right)^3{\text{ }} = {\text{ }}a^3{\text{ }} + {\text{ }}b^3{\text{ }} + {\text{ }}3a^{2}b{\text{ }} + {\text{ }}3a{b}^2{\text{ }} \]
\[ \Rightarrow {b^3} + {\text{ }}8{\text{ }} + {\text{ }}6{b^2} + {\text{ }}126{\text{ }}-{b^3} = {\text{ }}56\]
\[ \Rightarrow 6{b^2} + {\text{ }}12b{\text{ }}-{\text{ }}48{\text{ }} = {\text{ }}0\]
\[ \Rightarrow {b^2} + {\text{ }}2b{\text{ }}-{\text{ }}8{\text{ }} = {\text{ }}0\]_______III By factoring, we get
\[{{b}^{2}}+\text{ }4b\text{ }-\text{ }2b\text{ }-\text{ }8=\text{ }0\]
Taking common we get
⇒ (b+4)(b-2)
⇒ b = – 4, 2
Again put the value of b in equation III we get,
Hence value of a = b + 2
= 2 + 2 = 4
So the value of a is 4 and value of b is 2.

Note: Since in Question only positive integer is asked So, we neglect the negative value of a
i.e. a = b + 2
     = – 4 + 2
     = – 2
And negative value of b i.e -4