For every integer n≥1,(n3+11n)⋅(8n−14n+27) is divisible by 6 7 42 84

Hint : Since you probably missed some degrees in the expression $(8n-14n+27)$ , I can only complete part of the task.

Using mathematical induction, we prove that $(n^3+11n)$ is divisible by $6$ for $n\in\mathbb{Z}^+$ .

1 STEP: Induction basis

$n=1\to 1^3+11\cdot1=12=2\cdot6\,\,\,\vdots\,\,\,6$

2 STEP : Inductive guess

$\text{For all}\,\,\,k\le n\,\,\,(k^3+11k)\,\,\,\vdots\,\,\,6$

3 STEP : Inductive transition, it is necessary to prove that

$\text{For}\,\,\,k=n+1\to \left((n+1)^3+11(n+1)\right)\,\,\,\vdots\,\,\,6$

$(n+1)^3+11(n+1)=n^3+3n^2+3n+1+11n+11=\\[0.3cm] =(n^3+11n)+(3n^2+3n)+12=\\[0.3cm] =\underbrace{(n^3+11n)}_{\text{divided into 6 hypothesis}}+3n(n+1)+6\cdot 2$

It remains to understand why $3n(n+1)\,\,\,\vdots\,\,\,6$ .

1 case : $n=2k$

$3n(n+1)=3\cdot(2k)\cdot(2k+1)=6\cdot\left(k(2k+1)\right)\,\,\,\vdots\,\,\,6$

2 case : $n=2k+1$

$3n(n+1)=3\cdot(2k+1)\cdot(2k+1+1)=6\cdot\left((k+1)(2k+1)\right)\,\,\vdots\,\,6$

Conclusion,

$(n+1)^3+11(n+1)=(n^3+11n)+3n(n+1)+6\cdot2\,\,\vdots\,\,6$Hint : Since you probably missed some degrees in the expression 

$(8n-14n+27)$ , I can only complete part of the task.

Using mathematical induction, we prove that $(n^3+11n)$ is divisible by $6$ for $n\in\mathbb{Z}^+$ .

1 STEP: Induction basis

$n=1\to 1^3+11\cdot1=12=2\cdot6\,\,\,\vdots\,\,\,6$

2 STEP : Inductive guess

$\text{For all}\,\,\,k\le n\,\,\,(k^3+11k)\,\,\,\vdots\,\,\,6$

3 STEP : Inductive transition, it is necessary to prove that

$\text{For}\,\,\,k=n+1\to \left((n+1)^3+11(n+1)\right)\,\,\,\vdots\,\,\,6$

$(n+1)^3+11(n+1)=n^3+3n^2+3n+1+11n+11=\\[0.3cm] =(n^3+11n)+(3n^2+3n)+12=\\[0.3cm] =\underbrace{(n^3+11n)}_{\text{divided into 6 hypothesis}}+3n(n+1)+6\cdot 2$

It remains to understand why $3n(n+1)\,\,\,\vdots\,\,\,6$ .

1 case : $n=2k$

$3n(n+1)=3\cdot(2k)\cdot(2k+1)=6\cdot\left(k(2k+1)\right)\,\,\,\vdots\,\,\,6$

2 case : $n=2k+1$

$3n(n+1)=3\cdot(2k+1)\cdot(2k+1+1)=6\cdot\left((k+1)(2k+1)\right)\,\,\vdots\,\,6$

Conclusion,

$(n+1)^3+11(n+1)=(n^3+11n)+3n(n+1)+6\cdot2\,\,\vdots\,\,6$Hint : Since you probably missed some degrees in the expression 

$(8n-14n+27)$ , I can only complete part of the task.

Using mathematical induction, we prove that $(n^3+11n)$ is divisible by $6$ for $n\in\mathbb{Z}^+$ .

1 STEP: Induction basis

$n=1\to 1^3+11\cdot1=12=2\cdot6\,\,\,\vdots\,\,\,6$

2 STEP : Inductive guess

$\text{For all}\,\,\,k\le n\,\,\,(k^3+11k)\,\,\,\vdots\,\,\,6$

3 STEP : Inductive transition, it is necessary to prove that

$\text{For}\,\,\,k=n+1\to \left((n+1)^3+11(n+1)\right)\,\,\,\vdots\,\,\,6$

$(n+1)^3+11(n+1)=n^3+3n^2+3n+1+11n+11=\\[0.3cm] =(n^3+11n)+(3n^2+3n)+12=\\[0.3cm] =\underbrace{(n^3+11n)}_{\text{divided into 6 hypothesis}}+3n(n+1)+6\cdot 2$

It remains to understand why $3n(n+1)\,\,\,\vdots\,\,\,6$ .

1 case : $n=2k$

$3n(n+1)=3\cdot(2k)\cdot(2k+1)=6\cdot\left(k(2k+1)\right)\,\,\,\vdots\,\,\,6$

2 case : $n=2k+1$

$3n(n+1)=3\cdot(2k+1)\cdot(2k+1+1)=6\cdot\left((k+1)(2k+1)\right)\,\,\vdots\,\,6$

Conclusion,

$(n+1)^3+11(n+1)=(n^3+11n)+3n(n+1)+6\cdot2\,\,\vdots\,\,6$