Question

Find the prime factorisation of the following numbers:
${100}^3-{49}^3+{10}^3-{61}^3$

Answer 1

Let us consider $49-10+61=100$, therefore, the given problem can be rewritten as:

$\left[\right(49)+(-10)+(61){]}^3-(49{)}^3-(-10{)}^3-(61{)}^3$

We know the identity: $(a+b+c{)}^3-a^3-b^3-c^3=3(a+b\left)\right(b+c\left)\right(c+a)$

Using the above identity taking $a=49$, $b=-10$ and $c=61$, the equation

$\left[\right(49)+(-10)+(61){]}^3-(49{)}^3-(-10{)}^3-(61{)}^3$ can be factorised as follows:

$\left[\right(49)+(-10)+(61){]}^3-(49{)}^3-(-10{)}^3-(61{)}^3=3(49-10)(-10+61)(61+49)=3\times 39\times 51\times 110$

$=2\times (3\times 3\times 3)\times 5\times 11\times 13\times 17=2\times 3^3\times 5\times 11\times 13\times 17$

Hence, ${100}^3-{49}^3+{10}^3-{61}^3=2\times 3^3\times 5\times 11\times 13\times 17$