Question

Remove the Brackets​

Answer 1

$\underline{\underline{\sf{\maltese\:\:Question}}}}$

$\bigstar\:\:\:\sf{n\left[15-n\left(2n+5\right)\left(n-3\right)+n\left\{-15-7n\left(n-4\right)\right\}+18\right]}$

$\underline{\underline{\sf{\maltese\:\:Answer}}}}$

$\bf{\bullet\:\:\:-9n^4+29n^3+33n}$

$\underline{\underline{\sf{\maltese\:\:Calculations}}}}$

$\sf{n\left[15-n\left(2n+5\right)\left(n-3\right)+n\left\{-15-7n\left(n-4\right)\right\}+18\right]}$

$=\sf{n\left(15-n\left(2n+5\right)\left(n-3\right)+n\left(-15-7n\left(n-4\right)\right)+18\right)}$

$=\sf{n(15-2n^3+n^2+15n+n\left(-15-7n\left(n-4\right)\right)+18)}$

$\sf{=n(15-2n^3+n^2+15n-15n-7n^2\left(n-4\right)+18)$

$\sf{=n(15-2n^3+n^2+15n-15n-7n^3+28n^2+18)}$

$\sf{=n(-2n^3-7n^3+n^2+28n^2+15n-15n+15+18)}$

$\sf{=n(-2n^3-7n^3+29n^2+15n-15n+15+18)}$

$\sf{=n(-9n^3+29n^2+15n-15n+15+18)}$

$\sf{=n(-9n^3+29n^2+15+18)}$

$\sf{=n(-9n^3+29n^2+33)}$

$\sf{=n\left(-9n^3\right)+n\cdot \:29n^2+n\cdot \:33}$

$\sf{=-9n^3n+29n^2n+33n}$

$\bf{=-9n^4+29n^3+33n}$

Answer 2

Solution:-

$\rm \to \:n [15 - n(2n + 5)(n - 3) + n \{ - 15 - 7n(n - 4) \} + 18 ]$

Now using BODMAS concepts

$\rm \to \:n [15 - n(2n + 5)(n - 3) + n \{ - 15 - 7n^{2} + 28n \} + 18 ]$

$\rm \to \:n [15 - n(2n + 5)(n - 3) - 15 n- 7n^{3} + 28n ^{2} + 18 ]$

$\rm \to \:n [15 - (2n ^{2} + 5n)(n - 3) - 15 n- 7n^{3} + 28n ^{2} + 18 ]$

$\rm \to \: n [15 - (2 {n}^{3} - 6 {n}^{2} + 5 {n}^{2} - 15n) - 15n - 7 {n}^{3} + 28 {n}^{2} + 18 ]$

$\rm \to \: n [15 - 2 {n}^{3} + 6 {n}^{2} - 5 {n}^{2} + 15n - 15n - 7 {n}^{3} + 28 {n}^{2} + 18 ]$

$\rm \to \: n [33 - 9{n}^{3} + 29 {n}^{2} ]$

$\rm \to \: 33n - 9 {n}^{4} + 29 {n}^{3}$

BODMAS

=> B means Bracket

=> O means Of

=> D means Divide

=> M means Multiplication

=> A means Addition

=> S means Subtraction