Question

this is a question of prmo

Answer 1

Answer:

If 121=1+x1+x2+x3+x4 what is x ?

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We want to find the solutions of 1+x+x2+x3+x4=121 . This is equal to find the roots of P(x)=x+x2+x3+x4−120 . Using the we know that any factor of 120 could be a root of the polynomial. In particular, P(3)=3+9+27+81−120=0 .

Now using rule we have:

3111341121313940−1201200

Now we got P(x)=(x−3)(x3+4x2+13x+40)=(x−3)Q(x)

Let's propose the following change of variable for Q(x) : y−43=x then we have: Q(y)=y3+233y+74027 Then let's consider y=u+v

y3+233y+74027=u3+v3+74027+(u+v)(3uv+233)

And now we have:

u3+v3+74027=0

3uv+233=0

Therefore:

u6+74027u3−12167729=0

Which implies:

u3=−37027±16563−−−−−√9

u=16563−−−−−√9−37027−−−−−−−−−−−−−√3exp(i2πn3)

v=−16563−−−−−√9−37027−−−−−−−−−−−−−−√3exp(−i2πn3)

for n∈{0,1,2}

So the 4 solutions are:

x1=3

x2=16563−−−−−√9−37027−−−−−−−−−−−−−√3+−16563−−−−−√9−37027−−−−−−−−−−−−−−√3−43

x3=16563−−−−−√9−37027−−−−−−−−−−−−−√3(1+i3–√2)+−16563−−−−−√9−37027−−−−−−−−−−−−−−√3(1−i3–√2)−43

x4=16563−−−−−√9−37027−−−−−−−−−−−−−√3(1−i3–√2)+−16563−−−−−√9−37027−−−−−−−−−−−−−−√3(1+i3–√2)−43

Answer 2

Answer:

Step-by-step explanation:

Now using rule we have:

3111341121313940−1201200

Now we got P(x)=(x−3)(x3+4x2+13x+40)=(x−3)Q(x)

Let's propose the following change of variable for Q(x) : y−43=x then we have: Q(y)=y3+233y+74027 Then let's consider y=u+v

y3+233y+74027=u3+v3+74027+(u+v)(3uv+233)

And now we have:

u3+v3+74027=0

3uv+233=0

Therefore:

u6+74027u3−12167729=0

Which implies:

u3=−37027±16563−−−−−√9

u=16563−−−−−√9−37027−−−−−−−−−−−−−√3exp(i2πn3)

v=−16563−−−−−√9−37027−−−−−−−−−−−−−−√3exp(−i2πn3)

for n∈{0,1,2}

So the 4 solutions are:

x1=3

x2=16563−−−−−√9−37027−−−−−−−−−−−−−√3+−16563−−−−−√9−37027−−−−−−−−−−−−−−√3−43

x3=16563−−−−−√9−37027−−−−−−−−−−−−−√3(1+i3–√2)+−16563−−−−−√9−37027−−−−−−−−−−−−−−√3(1−i3–√2)−43

x4=16563−−−−−√9−37027−−−−−−−−−−−−−√3(1−i3–√2)+−16563−−−−−√9−37027−−−−−−−−−−−−−−√3(1+i3–√2)−43v