Question

Find the prime factoristion of the following numbers:
${30}^3-{12}^3-{10}^3-8^3$

Answer 1

Let us consider $12+10+8=30$, therefore, the given problem can be rewritten as:

$\left[\right(12)+(10)+(8){]}^3-(12{)}^3-(10{)}^3-(8{)}^3$

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 $a=12$, $b=10$ and $c=8$, the equation

$\left[\right(12)+(10)+(8){]}^3-(12{)}^3-(10{)}^3-(8{)}^3$ can be factorised as follows:

$\left[\right(12)+(10)+(8){]}^3-(12{)}^3-(10{)}^3-(8{)}^3=3(12+10)(10+8)(8+12)=3\times 22\times 18\times 20$

$=(2\times 2\times 2\times 2)\times (3\times 3\times 3)\times 5\times 11=2^4\times 3^3\times 5\times 11$

Hence, ${30}^3-{12}^3-{10}^3-8^3=2^4\times 3^3\times 5\times 11$