Question

Factorise: a6 -7a3-8

Answer 1

= > a^6 - 7a^3 - 8


Splitting the middle term ( i.e. 7 ) in such a manner so that the product of coeeficeints of first and last term will be equal to the product of splited parts of 7.

7 = 8 - 1 will satisfy the equation as the product of 8 and 1 is equal to the product of 1 and 8 .


= > a^6 - ( 8 - 1 )a^3 - 8

= > a^6 - 8a^3 + a^3 - 8

= > a^3( a^3 - 8 ) + ( a^3 - 8 )

= > ( a^3 - 8 )( a^3 + 1 )

= > ( a^3 - 2^3 )( a^3 + 1 )


From the properties of expansion, we know :
• a^3 - b^3 = ( a - b )( a^2 + ab + b^2 )
• a^3 + b^3 = ( a + b )( a^2 - ab + b^2 )


Therefore,

= > ( a - 2 )( a^2 + 2a + 4 )( a + 1 )( a^2 - a + 1 )

= > ( a - 2 )( a + 1 )( a^2 + 2a + 4 )( a^2 - a + 1 )



Hence,
a^6 - 7a^3 - 8 in factorized form is ( a - 2 )( a + 1 )( a^2 + 2a + 4 )( a^2 - a + 1 )

$\:$

Answer 2

$\underline{\mathfrak{\huge{Question:}}}$

Factorise the equation :- $a^{6} - 7a^{3} - 8$

$\underline{\mathfrak{\huge{Answer:}}}$

We can first simplify it by the splitting middle term method, which will be as :-

Finding such numbers which have their products equal to the extremes and their sum equal to the mean ( extremes means the end terms of the equation and mean means the middle term of the equation ).

Using the splitting middle term method, we get :-

=》 $a^{6} + a^{3} - 8a^{3} - 8$

=》 $a^{3} ( a^{3} + 1 ) - 8 ( a^{3} + 1 )$

=》 $( a^{3} - 8 ) ( a^{3} + 1 )$

=》 $( a^{3} - 2^{3} ) ( a^{3} + 1^{3} )$ ...(1)

We need to simplify further this obtained equation.

We can use the identities :-

⊙$x^{3} - y^{3} = ( x - y ) ( x^{2} + xy + y^{2} )$

Here, we can keep:-

x = a
y = 2

=》 $a^{3} - 2^{3} = ( a - 2 ) ( a^{2} + 2a + 4 )$ ...(2)

⊙$x^{3} + y^{3} = ( x + y ) ( x^{2} - xy + y^{2} )$

Here, we can keep:-

x = a
y = 1

=》 $a^{3} + 1^{3} = ( a + 1 ) ( a^{2} - a + 1 )$ ...(3)

We know that :-

(1) = (2) × (3)

=》 $a^{6} - 7a^{3} - 8 = ( a - 2 ) ( a^{2} + 2a + 4 ) ( a + 1 ) ( a^{2} - a + 1 )$

There's your factorised equation !