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System of equations II

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,753
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
Feb 7, 2012
2,725
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

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

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

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

Adding we get:

\(\displaystyle 6p+6q+6r+6s=48\)

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:

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

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

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

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

Adding, we get:

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

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. :D
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,753
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:

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

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

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

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

Adding we get:

\(\displaystyle 6p+6q+6r+6s=48\)

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:

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

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

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

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

Adding, we get:

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

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. :D
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. (Tongueout)

@Opalg, I really like your method, in fact, I like all of your posts so very much!:)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
To the corner with me...(Dull)

meincorner.jpg

(Smirk)
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,753

mente oscura

Well-known member
Nov 29, 2013
172
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. :eek:

I follow the succession (up):

[tex]4^3p+5^3q+6^3r+7^3s=7^3[/tex]

[tex]3^3p+4^3q+5^3r+6^3s=5^3[/tex]

[tex]2^3p+3^3q+4^3r+5^3s=3^3[/tex]

[tex]1^3p+2^3q+3^3r+4^3s=1^3[/tex]

[tex]0^3p+1^3q+2^3r+3^3s=-1^3[/tex]. (a)

[tex]-1^3p+0^3q+1^3r+2^3s=-3^3[/tex]. (b)

[tex]-2^3p-1^3q+0^3r+1^3s=-5^3[/tex]. (c)

[tex]-3^3p-2^3q-1^3r+0^3s=-7^3[/tex]. (d)

[tex]-4^3p-3^3q-2^3r-1^3s=-9^3[/tex]

Therefore:

[tex]4^3p+3^3q+2^3r+1^3s=+9^3=729[/tex]

Now:

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

[tex](b)-(c)=7p+q+r+7s=98[/tex]. (f)

[tex](f)-(e)=4p+4s=60 \rightarrow{}p+s=15[/tex]. (g)

For (e):

[tex]3p+q+r+3s=38[/tex]

[tex]45+q+r=38 \rightarrow{}q+r=-7[/tex]

Therefore:

[tex]p+s+q+r=15-7=8[/tex]


Regards.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,753
Hello.

I, as always, the more gross. :eek:

I follow the succession (up):

[tex]4^3p+5^3q+6^3r+7^3s=7^3[/tex]

[tex]3^3p+4^3q+5^3r+6^3s=5^3[/tex]

[tex]2^3p+3^3q+4^3r+5^3s=3^3[/tex]

[tex]1^3p+2^3q+3^3r+4^3s=1^3[/tex]

[tex]0^3p+1^3q+2^3r+3^3s=-1^3[/tex]. (a)

[tex]-1^3p+0^3q+1^3r+2^3s=-3^3[/tex]. (b)

[tex]-2^3p-1^3q+0^3r+1^3s=-5^3[/tex]. (c)

[tex]-3^3p-2^3q-1^3r+0^3s=-7^3[/tex]. (d)

[tex]-4^3p-3^3q-2^3r-1^3s=-9^3[/tex]

Therefore:

[tex]4^3p+3^3q+2^3r+1^3s=+9^3=729[/tex]

Now:

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

[tex](b)-(c)=7p+q+r+7s=98[/tex]. (f)

[tex](f)-(e)=4p+4s=60 \rightarrow{}p+s=15[/tex]. (g)

For (e):

[tex]3p+q+r+3s=38[/tex]

[tex]45+q+r=38 \rightarrow{}q+r=-7[/tex]

Therefore:

[tex]p+s+q+r=15-7=8[/tex]


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!:eek: