Solving A Cubic System #algebra #polynomials

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[SidneiMV]

a³ + b³ = 2√5
a²b + ab² = √5

a³ + b³ = 2√5
3a²b + 3ab² = 3√5
(a + b)³ = 5√5

a + b = √5
ab(a + b) = √5
ab = 1

t² - t√5 + 1 = 0
t = (√5 ± 1)/2

(a, b) = {[(√5 + 1)/2, (√5 - 1)/2]; [(√5 - 1)/2, (√5 + 1)/2]}

[nasrullahhusnan]

a³+b³=sqrt(5) (1)
a²b+ab²=sqrt(5) (2)
Note that LHS of the two given equation reminds us to
(a+b)³=a³+b³+3a²b+3ab²
=a³+b³+3(a²b+ab²)
=5sqrt(5)
=[sqrt(5)]³ --> a+b=sqrt(5)
From (2) ab(a+b)=sqrt(5)
absqrt(5)=sqrt(5)
ab=1
a[sqrt(5)-a]=1 -->
a²-asqrt(5)+1=0 -> a=½[sqrt(5)±1]
=±½[1±sqrt(5)]
=±ß
where ß is golden ratio
b=-1/ß or b=1/ß

[Qermaq]

The reason a + b can't be 0 is because then a = x and b = -x, and x^3 + (-x)^3 would have to be 0 also. But it's 2sqrt(5).

[walterwen]

Solving A Cubic System: a³ + b³ = 2√5, a²b + ab² = √5; x =?
a³ + b³ + 3(a²b + ab²) = 2√5 + 3√5, (a + b)³ = 5√5 = (√5)³, a + b = √5
a²b + ab² = ab(a + b) = √5(ab) = √5, ab = 1; (a – b)² = (a + b)² – 4ab = 5 – 4 = 1
a – b = ± 1, a + b = √5, 2a = √5 ± 1, a = (√5 ± 1)/2; b = (√5 –/+ 1)/2
Answer check:
a = (√5 ± 1)/2, b = (√5 –/+ 1)/2: a + b = √5, ab = 1
a³ + b³ = (a + b)[(a + b)² – 3ab] = √5(5 – 3) = 2√5; Confirmed
a²b + ab² = ab(a + b) = √5; Confirmed
Final answer:
a = (√5 + 1)/2, b = (√5 – 1)/2 or a = (√5 – 1)/2, b = (√5 + 1)/2

[uggupuggu]

a+b = sqrt(5)

ab(a+b) = sqrt(5)

ab= 1

it follows that a and b are phi and 1/phi