Question

Factorise $8{{x}^{3}}+27{{y}^{3}}+125{{z}^{3}}-90xyz$

Answer 1

Hint: In this problem we have an equation which is having coefficient for each variable .to Factorise this equation we will convert the coefficients into whole powers of preceding variables now the whole equation in the form of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ .In algebra we have formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$ by applying the above equation we will get factories of the given equation as $\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$

Formula used:
1. ${{a}^{m}}.{{b}^{m}}={{\left( ab \right)}^{m}}$
2. ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$

Complete step by step solution:
 Given that
$8{{x}^{3}}+27{{y}^{3}}+125{{z}^{3}}-90xyz$
Considering each term individually
Considering “$8{{x}^{3}}$:
We know that $8={{2}^{3}}$ ,substituting this value in $8{{x}^{3}}$,then we will get
$8{{x}^{3}}={{2}^{3}}\left( {{x}^{3}} \right)$
Using the formula ${{a}^{m}}.{{b}^{m}}={{\left( ab \right)}^{m}}$, then will get
$\Rightarrow 8{{x}^{3}}={{\left( 2x \right)}^{3}}$
Considering $27{{y}^{3}}$ :
We know that $27={{3}^{3}}$ ,substituting this value in $27{{y}^{3}}$,then we will get
$27{{y}^{3}}={{3}^{3}}\left( {{y}^{3}} \right)$
Using the formula ${{a}^{m}}.{{b}^{m}}={{\left( ab \right)}^{m}}$, then will get
$\Rightarrow 27{{y}^{3}}={{\left( 3y \right)}^{3}}$
Considering $125{{z}^{3}}$ :
We know that $125={{5}^{3}}$ ,substituting this value in $125{{z}^{3}}$,then we will get
$125{{z}^{3}}={{5}^{3}}\left( {{z}^{3}} \right)$
Using the formula ${{a}^{m}}.{{b}^{m}}={{\left( ab \right)}^{m}}$, then will get
$\Rightarrow 125{{z}^{3}}={{\left( 5z \right)}^{3}}$
Considering $90xyz$ :
We know that $90=2\times 3\times 3\times 5$ ,substituting this value in $90xyz$,then we will get
$90xyz=2\times 3\times 3\times 5\times x\times y\times z$
Rearranging the above terms, then above equation is modified as
$90xyz=3\times \left( 2x \right)\times \left( 3y \right)\times \left( 5z \right)$
Substituting the all the values in the given equation
$8{{x}^{3}}+27{{y}^{3}}+125{{z}^{3}}-90xyz={{\left( 2x \right)}^{3}}+{{\left( 3y \right)}^{3}}+{{\left( 5z \right)}^{3}}-3\left( 2x \right)\left( 3y \right)\left( 5z \right)$
Comparing the above equation with ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ then the values of $a,$ $b,$ $c,$ are
$a=2x$
$b=3y$
$c=5z$
Applying the formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$ then we will get
${{\left( 2x \right)}^{3}}+{{\left( 3y \right)}^{3}}+{{\left( 5z \right)}^{3}}-3\left( 2x \right)\left( 3y \right)\left( 5z \right)=\left( 2x+3y+5z \right)\left( {{\left( 2x \right)}^{2}}+{{\left( 3y \right)}^{2}}+{{\left( 5z \right)}^{2}}+\left( 3\times 2\times x\times y \right)-\left( 3\times 5\times y\times z \right)-\left( 5\times 2\times x\times z \right) \right)$
Expanding and simplifying the terms in the above equation, then we will get
${{\left( 2x \right)}^{3}}+{{\left( 3y \right)}^{3}}+{{\left( 5z \right)}^{3}}-3\left( 2x \right)\left( 3y \right)\left( 5z \right)=\left( 2x+3y+5z \right)\left( 4{{x}^{2}}+9{{y}^{2}}+25{{z}^{2}}+6xy-15yz-10xz \right)$

Note: There is no need to consider each term individually if you are comfort with the exponential you directly go to the step where we have applied the formula ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$