Question

Factorise: $ 8{{x}^{6}}+95{{x}^{3}}+1 $
A. $ \left( {{x}^{2}}+5x+1 \right)\left( 4{{x}^{4}}-10{{x}^{3}}+23{{x}^{3}}-5x+1 \right) $
B. $ \left( 2{{x}^{2}}+5x+1 \right)\left( 4{{x}^{4}}-10{{x}^{3}}+23{{x}^{2}}-5x+1 \right) $
C. $ \left( 2{{x}^{2}}+x+1 \right)\left( 4{{x}^{4}}-10{{x}^{3}}+23{{x}^{3}}-5x+1 \right) $
D. $ \left( 2{{x}^{2}}+5x+1 \right)\left( 4{{x}^{4}}-10{{x}^{2}}+23{{x}^{3}}-5x+1 \right) $

Answer 1

Hint: We use the identity form of $ {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right) $ to factor the given equation of $ 8{{x}^{6}}+95{{x}^{3}}+1 $ . We assume the values for $ a=2{{x}^{2}},b=5x,c=1 $ . We form the required identity and find the factorisation.

Complete step by step solution:
To Factorise $ 8{{x}^{6}}+95{{x}^{3}}+1 $ , we take the help of the identity
 $ {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right) $ .
We first need to convert the equation $ 8{{x}^{6}}+95{{x}^{3}}+1 $ to the form of $ {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc $ .
So, $ 8{{x}^{6}}+95{{x}^{3}}+1=8{{x}^{6}}+125{{x}^{3}}+1-30{{x}^{3}} $ .
Equating the above formation with $ {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc $ , we get $ a=2{{x}^{2}},b=5x,c=1 $ .
Therefore, automatically the form of $ -3abc=-3\times \left( 2{{x}^{2}} \right)\times \left( 5x \right)\times \left( 1 \right)=-30{{x}^{3}} $ .
Therefore, $ 8{{x}^{6}}+95{{x}^{3}}+1={{\left( 2{{x}^{2}} \right)}^{3}}+{{\left( 5x \right)}^{3}}+{{1}^{3}}-3\times \left( 2{{x}^{2}} \right)\times \left( 5x \right)\times \left( 1 \right) $ .
Using the identity form we get
 $
   8{{x}^{6}}+95{{x}^{3}}+1 \\
  ={{\left( 2{{x}^{2}} \right)}^{3}}+{{\left( 5x \right)}^{3}}+{{1}^{3}}-3\times \left( 2{{x}^{2}} \right)\times \left( 5x \right)\times \left( 1 \right) \\
  =\left( 2{{x}^{2}}+5x+1 \right)\left\{ {{\left( 2{{x}^{2}} \right)}^{2}}+{{\left( 5x \right)}^{2}}+{{1}^{2}}-\left( 2{{x}^{2}} \right)\left( 5x \right)-\left( 5x \right)\times 1-\left( 2{{x}^{2}} \right)\times 1 \right\} \\
  =\left( 2{{x}^{2}}+5x+1 \right)\left( 4{{x}^{4}}+25{{x}^{2}}+1-10{{x}^{3}}-5x-2{{x}^{2}} \right) \;
  =\left( 2{{x}^{2}}+5x+1 \right)\left( 4{{x}^{4}}-10{{x}^{3}}+23{{x}^{2}}-5x+1 \right) \;
 $
The correct option is B.
So, the correct answer is “Option B”.

Note: We try to show that the factorisation is well defined. We assume the values $ x=1 $ for both $ 8{{x}^{6}}+95{{x}^{3}}+1=\left( 2{{x}^{2}}+5x+1 \right)\left( 4{{x}^{4}}-10{{x}^{3}}+23{{x}^{2}}-5x+1 \right) $ .
Putting the value, we get $ 8{{x}^{6}}+95{{x}^{3}}+1=8\times {{1}^{6}}+95\times {{1}^{3}}+1=8+95+1=104 $ and
 $
   \left( 2{{x}^{2}}+5x+1 \right)\left( 4{{x}^{4}}-10{{x}^{3}}+23{{x}^{2}}-5x+1 \right) \\
  =\left( 2\times {{1}^{2}}+5\times 1+1 \right)\left( 4\times {{1}^{4}}-10\times {{1}^{3}}+23\times {{1}^{2}}-5\times 1+1 \right) \\
  =\left( 8 \right)\left( 13 \right) \;
  =104 \;
 $
Therefore, the factorisation $ 8{{x}^{6}}+95{{x}^{3}}+1=\left( 2{{x}^{2}}+5x+1 \right)\left( 4{{x}^{4}}-10{{x}^{3}}+23{{x}^{2}}-5x+1 \right) $ is defined.