E times pond[2a+3b+4c]^(2)

Subtract (2a - 3b + 4c) from the sum of (a + 3b - 4c), (4a - b +9c) and (-2b + 3c -a )

Solve : ( a^(-2) times b^(-3) times c^2 )/ ( b^-4 times c^-3 )

If (a)/(b)=(c)/(d)=(e)/(f)=3 , then (2a^2 + 3c^(2) + 4e^(2))/(2b^(2) + 3d^(2) + 4f^(2))=?

"(a^(-3)times b^(-2)times c^(-5)times d^(5))/(a^(4)times b times c^(-3)times d^(4))

Factorise : 2ab^(2) c - 2a + 3b^(3) c - 3b - 4b^(2) c^(2) + 4c

Evaluate : a^2 times a^4 times b^5 times c^2 times b^2 times c^5

Simplify : (Use formulae) (2a-3b+4c)^2-(2a+3b-4c)^2

Write these as a single power? a) (4^2)^3" " b) (3^3)^2 xx 9^4 " " c) ((1/2)^3)^4" " d) (2^3)^4 xx 2^6" "