# `(r + 3s)^6` Use the Binomial Theorem to expand and simplify the expression.

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A simple method used to solve this binomial is the use of pascal's triangle:

The expansion of our example is expanded as follows using pascal's triangle:

Now let's use the above expansion to solve our problem:

` ` `(r+3s)^6 = 1*r^6 + 6*r^5 * (3s)^1 + 15* r^4 * (3s)^2 + 20* r^3 * (3s)^3 + 15 * r^2 * (3s)^4 + 6*r*(3s)^5 + 1* (3s)^6`

Now we can evaluate our answer as follows:

`(r+3s)^6 = r^6 + 6*3 * r^5 *s^1 + 15 * 9 * r^4 * s^2 +20 * 27 * r^3 * s^3 + 15 * 81 * r^2 * s^4 + 6 * 3^5 * r * s^5 + 729 * s^6`

`(r+3s)^6 = r^6 + 18 * r^5 * s + 135 * r^4 * s^2 + 540*r^3*s^3 + 1215 * r^2 * s^4 + 1458 *r *s^5 + 729 * s^6`

Above is the final answer.

We have to use the binomial theorem to expand the expression, so use the formula:

`(a+b)^n=sum_(k=0)^n((n),(k))a^(n-k)b^k`

`:.(r+3s)^6=((6),(0))r^(6-0)(3s)^0+((6),(1))r^(6-1)(3s)^1+((6),(2))r^(6-2)(3s)^2+((6),(3))r^(6-3)(3s)^3+((6),(4))r^(6-4)(3s)^4+((6),(5))r^(6-5)(3s)^5+((6),(6))r^(6-6)(3s)^6`

`=r^6+(6!)/(1!(6-1)!)r^5(3s)+(6!)/(2!(6-2)!)r^4(9s^2)+(6!)/(3!(6-3)!)r^3(27s^3)+(6!)/(4!(6-4)!)r^2(81s^4)+(6!)/(5!(6-5)!)r^1(243s^5)+729s^6`

`=r^6+(6*5!)/(5!)(3r^5s)+(6*5*4!)/(2*4!)(9r^4s^2)+(6*5*4*3!)/(3!*3*2*1)(27r^3s^3)+(6*5*4!)/(4!*2*1)(81r^2s^4)+(6*5!)/(5!)(243rs^5+729s^6`

`=r^6+18r^5s+135r^4s^2+540r^3s^3+1215r^2s^4+1458rs^5+729s^6`