Question

Factorise each of the following equation:
I. $ 27{y^3} + 125{z^3} $
II. $ 64{m^3} - 343{n^3} $

Answer 1

Hint: In these types of equations we need to see that the values can be converted to cube or square of any number and then put them in the formula to get the factorized value.
 $ ({a^3} + {b^{^3}}) = (a + b)({a^2} - ab + {b^2}) $
 $ ({a^3} - {b^{^3}}) = (a - b)({a^2} + ab + {b^2}) $

Complete step-by-step answer:
Equation we are having to factorise is:
 $ 27{y^3} + 125{z^3} $
We can write the equation in the form of cubes as below
 $ {3^3}{y^3} + {5^3}{z^3} \Rightarrow {(3y)^3} + {(5z)^3} $
We have to consider $ '3y' $ as ‘a’ and $ '5z' $ as ‘b’ and then putting them in the formula given below and solving it:
 $ ({a^3} + {b^{^3}}) = (a + b)({a^2} - ab + {b^2}) $
 $\Rightarrow ({(3y)^3} + {(5z)^{^3}}) = (3y + 5z)({(3y)^2} - (3y)(5z) + {(5z)^2}) $
 $ = (3y + 5z)(9{y^2} - 15yz + 25{z^2}) $
Hence, Factorized value of $ 27{y^3} + 125{z^3} $ is $ (3y + 5z)(9{y^2} - 15yz + 25{z^{2)}} $
Similarly,
Equation we are having to factorise is:
  $ 64{m^3} - 343{n^3} $
We can write the equation in form of cubes as below
 $ {4^3}{m^3} - {7^3}{n^3} \Rightarrow {(4m)^3} - {(7n)^3} $
We have to consider $ '4m' $ as ‘a’ and $ '7n' $ as ‘b’ and then putting them in the formula given below and solving it:
 $ ({a^3} - {b^{^3}}) = (a - b)({a^2} + ab + {b^2}) $
 $\Rightarrow ({(4m)^3} - {(7n)^{^3}}) = (4m - 7n)({(4m)^2} + (4m)(7n) + {(7n)^2}) $
 $ = (4m - 7n)(16{m^2} + 28mn + 49{n^2}) $
Hence, Factorized value of $ 64{m^3} - 343{n^3} $ is $ (4m - 7n)(16{m^2} + 28mn + 49{n^2}) $

Note: A simple trick can be used to remember the two formulas, as we can see on the L.H.S of the equation its either $ ({a^3} - {b^3}) $ or $ ({a^3} + {b^3}) $ so accordingly on the first part of the R.H.S of the equation i.e. $ (a - b) $ or $ (a + b) $ similar – or + sign will be present and opposite – or + sign will be present on the second part of the R.H.S of the equation i.e. $ (ab) $ .