Question
Factorise :$a^6- 7a^3- 8$
✓
Answer
We know that,
$a^3 + b^3 = ( a + b )( a^2 - ab + b^2 ) ....(1)$
$a^3 - b^3 = ( a - b )( a^2 + ab + b^2 ) ....(2)$
$a^6- 7a^3 - 8$
$= a^6 - 8a^3 + a^3 - 8$
$= a^3( a^3 - 8) + 1( a^3 - 8 )$
$= ( a^3 + 1 )( a^3 - 8 )$
$= ( a^3 + 1^3 )( a^3 - 2^3 )$
$= ( a + 1 )( a^2 - a + 1 )( a - 2 )( a^2 + 2a + 4 )$
$[$ From$(1)$ and $(2) ]$
$= ( a + 1 )( a - 2)( a^2 - a + 1 )( a^2 + 2a + 4 )$
$a^3 + b^3 = ( a + b )( a^2 - ab + b^2 ) ....(1)$
$a^3 - b^3 = ( a - b )( a^2 + ab + b^2 ) ....(2)$
$a^6- 7a^3 - 8$
$= a^6 - 8a^3 + a^3 - 8$
$= a^3( a^3 - 8) + 1( a^3 - 8 )$
$= ( a^3 + 1 )( a^3 - 8 )$
$= ( a^3 + 1^3 )( a^3 - 2^3 )$
$= ( a + 1 )( a^2 - a + 1 )( a - 2 )( a^2 + 2a + 4 )$
$[$ From$(1)$ and $(2) ]$
$= ( a + 1 )( a - 2)( a^2 - a + 1 )( a^2 + 2a + 4 )$
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