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Factorise: $8{a^3} + 125{b^3} - 64{c^3} + 120abc$

Last updated date: 22nd Mar 2023
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Hint- Here, we will be using the formula for $\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$.

The given expression which needs to get factorised as $8{a^3} + 125{b^3} - 64{c^3} + 120abc$
As we know that ${x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right){\text{ }} \to {\text{(1)}}$
The given expression can be represented in the form of LHS of above equation as
$8{a^3} + 125{b^3} - 64{c^3} + 120abc = {\left( {2a} \right)^3} + {\left( {5b} \right)^3} + {\left( { - 4c} \right)^3} - 3\left( {2a} \right)\left( {5b} \right)\left( { - 4c} \right)$
Now replace $2a$ by $x$, $5b$ by $y$ and $- 4c$ by $z$, we get
$8{a^3} + 125{b^3} - 64{c^3} + 120abc = {x^3} + {y^3} + {z^3} - 3xyz$
Using equation (1), the above can be written as
$8{a^3} + 125{b^3} - 64{c^3} + 120abc = {x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)$
Now, let us back substitute the values of $x$, $y$ and $z$ in the above expression
$8{a^3} + 125{b^3} - 64{c^3} + 120abc = \left( {2a + 5b - 4c} \right)\left[ {{{\left( {2a} \right)}^2} + {{\left( {5b} \right)}^2} + {{\left( { - 4c} \right)}^2} - \left( {2a} \right)\left( {5b} \right) - \left( {5b} \right)\left( { - 4c} \right) - \left( { - 4c} \right)\left( {2a} \right)} \right] \\ \Rightarrow 8{a^3} + 125{b^3} - 64{c^3} + 120abc = \left( {2a + 5b - 4c} \right)\left( {4{a^2} + 25{b^2} + 16{c^2} - 10ab + 20bc + 8ca} \right) \\$
Therefore, as shown above the given expression is factorised into two factors.

Note- In these types of problems, we convert the given expression in a form which can be easily simplified with a help of a known formula. Here, we converted it in the form of $\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)$ and then its formula is used for factorisation.