Find the area of the region bounded by the curve `y-3` and the lines `y = x + 6` and `y =0`.

Explanation

We have, `y=x^{3}, y=x+6` and `x=0`



`therefore {x}^{3}={x}+6 `


`Rightarrow {x}^{3}-{x}=6`


`Rightarrow {x}^{3}-{x}-6=0`


`Rightarrow {x}^{2}({x}-2)+2 {x}({x}-2)+3({x}-2)=0`


`Rightarrow({x}-2)({x}^{2}+2 {x}+3)=0`



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