The factors of `8a^3+b^3-6a b+1` are
(a) `(2a+b-1)(4a^2+b^2+1-3a b-2a)`
(b)`(2a-b+1)(4a^2+b^2-4a b+1-2a+b)`
(c)`(2a+b+1)(4a^2+b^2+1-2a b-b-2a)`
(d)`(2a-1+b)(4a^2+1-4a-b-2a b)`

Show that: |1+a^2-b^2 2a b-2b2a b1-a^2+b^2 2a2b-2a1-a^2-b^2|=(1+a^2+b^2)^3

Simplify: a^2b(a-b^2)+a b^2(4a b-2a^2)-a^3b(1-2b)

Simplify: a^2b(a^3-a+1)-a b(a^4-2a^2+2a)-b(a^3-a^2-1)

Simplify: 4a b(a-b)-6a^2(b-b^2)-3b^2(2a^2-a)+2a b(b-a)

Simplify: -1/2a^2b^2c+1/3a b^2c-1/4a b c^2-1/5c b^2a^2+1/6c b^2a-1/7c^2a b+1/8c a^2bdot

If x+2 is a factor of x^2+a x+2b and a+b=4 , then (a) a=1,\ b=3 (b) a=3,\ b=1 (c) a=-1,\ b=5 (d) a=5,\ b=-1

If a ,b ,a n dc are in H.P., then th value of ((a c+a b-b c)(a b+b c-a c))/((a b c)^2) is ((a+c)(3a-c))/(4a^2c^2) b. 2/(b c)-1/(b^2) c. 2/(b c)-1/(a^2) d. ((a-c)(3a+c))/(4a^2c^2)

If a+ b = 4 and ab =3 , find (1)/(b^(2)) + (1)/(a^(2))

If a statement is true for all the values of the variable, such statements are called as identities. Some basic identities are : (1) (a+b)^(2)=a^(2)+2ab+b^(2)=(a-b)^(2)+4ab (3) a^(2)-b^(2)=(a+b)(a-b) (4) (a+b)^(3)=a^(3)+b^(3)+3ab(a+b) (6) a^(3)+b^(3)=(a+b)^(3)=3ab(a+b)=(a+b) (a^(2)-ab) (8) (a+b+c)^(2)=a^(2)+b^(2)+c^(2)+2ab+2bc+2ca=a^(2)+b^(2)+c^(2)+2abc((1)/(a)+(1)/(b)+(1)/(c)) (10) a^(3)+b^(3)+c^(3)-3abc=(a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ca) =1/2(a+b+c)[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)] If a+b+c=0,thena^(3)+b^(3)+c^(3)=3abc If x,y, z are different real umbers and (1)/((x-y)^(2))+(1)/((y-z)^(2))+(1)/((z-x)^(2))=((1)/(x-y)+(1)/(y-z)+(1)/(z-x))^2+lamda then the value of lamda is

Find the value of 2a^(3)-b^(4)+3a^(2)b^(3) when a =3 , b=-1 .