# System of equations II

#### anemone

##### MHB POTW Director
Staff member
Consider the system of equations:

$p+8q+27r+64s=1$

$8p+27q+64r+125s=27$

$27p+64q+125r+216s=125$

$64p+125q+216r+343s=343$

Evaluate $p+q+r+s$ and $64p+27q+8r+s$.

#### Opalg

##### MHB Oldtimer
Staff member
Let \begin{aligned}a_n &= pn^3 + q(n+1)^3 + r(n+2)^3 + s(n+3)^3 \\ &= (p+q+r+s)n^3 + 3(q+2r+3s)n^2 + 3(q+4r+9s)n + (q+8r+27s).\qquad(*) \end{aligned} We are told that $a_1=1^3$, $a_2 = 3^3$, $a_3 = 5^3$ and $a_4 = 7^3$. So for those four values of $n$, $a_n = (2n-1)^3$. But a cubic polynomial is determined by its values at four points, and therefore $$a_n = (2n-1)^3\qquad(**)$$ for all $n$.

Compare the coefficients of $n^3$ in (*) and (**) to get $p+q+r+s = 8$.

Then put $n=-4$ in (*) and (**) to get $(-4)^3p + (-3)^3q + (-2)^3r + (-1)^3s = (-9)^3$, or $64p+27q+8r+s = 729.$

#### MarkFL

Staff member
Here is my solution:

a) Evaluate $$\displaystyle p+q+r+s$$

Multiply the first equation by -1, the second by 3 the third by -3 and the system becomes:

$$\displaystyle -p-8q-27r-64s=-1$$

$$\displaystyle 24p+81q+192r+375s=81$$

$$\displaystyle -81p-192q-375r-648s=-375$$

$$\displaystyle 64p+125q+216r+343s=343$$

$$\displaystyle 6p+6q+6r+6s=48$$

Hence:

$$\displaystyle p+q+r+s=8$$

b) Evaluate $$\displaystyle 64p+27q+8r+s$$

Multiply the first equation by -56, the second by 140 the third by -120 and the fourth by 35 and the system becomes:

$$\displaystyle -56p-448q-1512r-3584s=-56$$

$$\displaystyle 1120p+3780q+8960r+17500s=3780$$

$$\displaystyle -3240p-7680q-15000r-25920s=-15000$$

$$\displaystyle 2240p+4375q+7560r+12005s=12005$$

$$\displaystyle 64p+27q+8r+s=729$$

I simply solved another 4X4 system (using a CAS) to determine the values by which to multiply the given equations. Opalg's solution is much more elegant. #### anemone

##### MHB POTW Director
Staff member
Let \begin{aligned}a_n &= pn^3 + q(n+1)^3 + r(n+2)^3 + s(n+3)^3 \\ &= (p+q+r+s)n^3 + 3(q+2r+3s)n^2 + 3(q+4r+9s)n + (q+8r+27s).\qquad(*) \end{aligned} We are told that $a_1=1^3$, $a_2 = 3^3$, $a_3 = 5^3$ and $a_4 = 7^3$. So for those four values of $n$, $a_n = (2n-1)^3$. But a cubic polynomial is determined by its values at four points, and therefore $$a_n = (2n-1)^3\qquad(**)$$ for all $n$.

Compare the coefficients of $n^3$ in (*) and (**) to get $p+q+r+s = 8$.

Then put $n=-4$ in (*) and (**) to get $(-4)^3p + (-3)^3q + (-2)^3r + (-1)^3s = (-9)^3$, or $64p+27q+8r+s = 729.$
Here is my solution:

a) Evaluate $$\displaystyle p+q+r+s$$

Multiply the first equation by -1, the second by 3 the third by -3 and the system becomes:

$$\displaystyle -p-8q-27r-64s=-1$$

$$\displaystyle 24p+81q+192r+375s=81$$

$$\displaystyle -81p-192q-375r-648s=-375$$

$$\displaystyle 64p+125q+216r+343s=343$$

$$\displaystyle 6p+6q+6r+6s=48$$

Hence:

$$\displaystyle p+q+r+s=8$$

b) Evaluate $$\displaystyle 64p+27q+8r+s$$

Multiply the first equation by -56, the second by 140 the third by -120 and the fourth by 35 and the system becomes:

$$\displaystyle -56p-448q-1512r-3584s=-56$$

$$\displaystyle 1120p+3780q+8960r+17500s=3780$$

$$\displaystyle -3240p-7680q-15000r-25920s=-15000$$

$$\displaystyle 2240p+4375q+7560r+12005s=12005$$

$$\displaystyle 64p+27q+8r+s=729$$

I simply solved another 4X4 system (using a CAS) to determine the values by which to multiply the given equations. Opalg's solution is much more elegant. Thank you for participating to both of you! Both are genius and perfect solutions to me, honestly speaking, albeit MarkFL's has cheated a bit in his method. @Opalg, I really like your method, in fact, I like all of your posts so very much! Staff member

#### anemone

##### MHB POTW Director
Staff member
To the corner with me... View attachment 1982 Please come back and don't stand(or sit?) in the corner, my sweetest admin! I can overlook this if... if you cook a nice meal for me! #### mente oscura

##### Well-known member
Consider the system of equations:

$p+8q+27r+64s=1$

$8p+27q+64r+125s=27$

$27p+64q+125r+216s=125$

$64p+125q+216r+343s=343$

Evaluate $p+q+r+s$ and $64p+27q+8r+s$.
Hello.

I, as always, the more gross. I follow the succession :

$$4^3p+5^3q+6^3r+7^3s=7^3$$

$$3^3p+4^3q+5^3r+6^3s=5^3$$

$$2^3p+3^3q+4^3r+5^3s=3^3$$

$$1^3p+2^3q+3^3r+4^3s=1^3$$

$$0^3p+1^3q+2^3r+3^3s=-1^3$$. (a)

$$-1^3p+0^3q+1^3r+2^3s=-3^3$$. (b)

$$-2^3p-1^3q+0^3r+1^3s=-5^3$$. (c)

$$-3^3p-2^3q-1^3r+0^3s=-7^3$$. (d)

$$-4^3p-3^3q-2^3r-1^3s=-9^3$$

Therefore:

$$4^3p+3^3q+2^3r+1^3s=+9^3=729$$

Now:

$$(a)-(d)=27p+9q+9r+27s=342 \rightarrow{}3p+q+r+3s=38$$. (e)

$$(b)-(c)=7p+q+r+7s=98$$. (f)

$$(f)-(e)=4p+4s=60 \rightarrow{}p+s=15$$. (g)

For (e):

$$3p+q+r+3s=38$$

$$45+q+r=38 \rightarrow{}q+r=-7$$

Therefore:

$$p+s+q+r=15-7=8$$

Regards.

#### anemone

##### MHB POTW Director
Staff member
Hello.

I, as always, the more gross. I follow the succession :

$$4^3p+5^3q+6^3r+7^3s=7^3$$

$$3^3p+4^3q+5^3r+6^3s=5^3$$

$$2^3p+3^3q+4^3r+5^3s=3^3$$

$$1^3p+2^3q+3^3r+4^3s=1^3$$

$$0^3p+1^3q+2^3r+3^3s=-1^3$$. (a)

$$-1^3p+0^3q+1^3r+2^3s=-3^3$$. (b)

$$-2^3p-1^3q+0^3r+1^3s=-5^3$$. (c)

$$-3^3p-2^3q-1^3r+0^3s=-7^3$$. (d)

$$-4^3p-3^3q-2^3r-1^3s=-9^3$$

Therefore:

$$4^3p+3^3q+2^3r+1^3s=+9^3=729$$

Now:

$$(a)-(d)=27p+9q+9r+27s=342 \rightarrow{}3p+q+r+3s=38$$. (e)

$$(b)-(c)=7p+q+r+7s=98$$. (f)

$$(f)-(e)=4p+4s=60 \rightarrow{}p+s=15$$. (g)

For (e):

$$3p+q+r+3s=38$$

$$45+q+r=38 \rightarrow{}q+r=-7$$

Therefore:

$$p+s+q+r=15-7=8$$

Regards.
Hey mente oscura, thanks for participating and you know what, my method is exactly the same as yours! I say our method is a good one too! 