Question

Factorise: 18a²bc – 12ab²c + 6abc² factor nikal ke bhatiya please aapko direct solution nahin likhna hai​

Answer 1

Answer:

3abc(6a-4b+2c)

Step-by-step explanation:

18a²bc – 12ab²c + 6abc²

3 x 6 x a²x b x c - 3 x 4 x a x b² x c + 3 x 2 x a x b x c²

{I'v made the single equation bold, because the equation is broken up.}

{Now we have to see which of the constants and variables are common.}

(I will underline the common ones.)

3 x 6 x a²x b x c - 3 x 4 x a x b² x c + 3 x 2 x a x b x c²

[The variables a, b and c are common because even if we have powers/indices on the variables, the base of the variables is same/common, i.e. a, b and c.]

So we solve,

3abc(6a - 4b + 2c)

{If we multiply this we get,}

18a²bc – 12ab²c + 6abc²