`M^4 + m^2 +1` Factorise.​

Find the L.C.M. of 6 and 9 by prime factorisation method.

The value of sum_(m=1)^(n) "tan"^(-1)((2m)/(m^(4) + m^(2) +2 )) is :

Factorise : x^(4)-1

Factorise m^(2)+m-56

If m+ (1)/(m-2)=4 , find m^(2) + m+1 =?

If m+ (1)/(m+2)=-4 , find m^(2) + m+1 =?

If m+ (1)/(m-2)=0 , find m^(5) + m^(4) + m^(3) + m^(2)+ m+1 = ?

If m+ (1)/(m+2)= 0 find m^(4) + m^(3) + m^(2) + m+1 =?

Evaluate : ( m^( 1/2 ) times m^( 1/4 ) times m^( 1/4 ) )/ ( m^(1/4 ) times m^( 1/4 ) times 1/( m^(1/16 )) )