IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

If a,b and c are the roots of x³+3x+3 = 0, then find the value of

a⁵+b⁵+c⁵

iberis22

1. a + b + c = 0

2. ab + bc + ca = 3

3. abc = -3

4. Since, a, b, c are the roots of equation,

a³ + 3a + 3 = 0

b³ + 3b + 3 = 0

c³ + 3c + 3 = 0

Thus, a³+ b³ + c³ + 3(a+b+c) + 9 = 0

But, a+b+c=0

Thus, a³ + b³ + c³ = -9

5. (a+b+c)² = a² + b² + c² + 2(ab + bc+ ca)

0 = a² + b² + c² + 2(3)

Thus, a² + b² + c² = -6

Now,

( a² + b² + c² )( a³ + b³ + c³ )

= a⁵ + b⁵ + c⁵ + a²b²(a+b) + b²c²(b+c) + a²c²(a+c)

putting a+b = -c

b+c = -a

c+a = -b

= a⁵ + b⁵ + c⁵ - abc(ab + bc + ca)

= a⁵ + b⁵ + c⁵ - (-3)(3)

LHS = (-6)(-9) = 54

Thus, a⁵ + b⁵ + c⁵ = 54-9 = 45

sboosy

x³+3x+3 = 0

a+b+c = 0

ab+bc+ca = 3

abc = -3

since a+b+c = 0

a³+b³+c³ = 3abc = -9

a²+b²+c² = (a+b+c)² - 2 [ab+bc+ca] = -6

(a³+b³+c³)(a²+b²+c²) = 54

= a⁵+b⁵+c⁵ + a³(b²+c²) + b³(a²+c²) + c³(a²+b²)

writing

b²+c² as (b+c)² - 2bc = a²-2bc

we get

2[a⁵+b⁵+c⁵] -2abc[a²+b²+c²] = 54

so

a⁵+b⁵+c⁵ = 45

pls correct me if wrong

man111

at first a₀s₁+a₁= 0 here s₁ is sum of root meas x₁+x₂+x₃ad s2 meas x₁²+x₂²+x₃²so here a₀=1 a₁=0,a₂=3,a₃=3 first we will calculate s₁

s₁=0, similarly a₀s₂+a₁s₁+2a₂= 0 so s₂+6=0, so s₂= - 6 similarly

a₀s₃+a₁s₂+a₂s₁+3a₃= 0 so s₃= - 9 for calculatig s₄we multiply this equatio

x. we get x⁴+3x²+3x=0

s₄+3s₂+3s=0 we get s₄+3(-6)+0=0 we get s₄=18

similarly multiply eq x²we get x⁵+3x³+3x²=0 we get s₅+3s₃+3s₂= 0 we get

s₅+3(-9)+3(-6)=0

s₅=27+18 = 45

hsbhatt

Just another method:

Since a³+3a+3 = 0 etc.,

We have a⁵+3a³+3a²=0 etc.(multiplying by a²)

Hence

a⁵+3

a³+3

a² = 0

Also

a³ = 3abc = -9 (Since

a=0)

Now

a² = (a+b+c)² - 2

ab = -6

Hence

a⁵ = 3*9+3*6 = 45

shreyaarya

to an extent same methods as used above but i have just used a conditional identity which can be proved easily using the above methods of multiplication of

( a² + b² + c² ).( a³ + b³ + c³ )etc..

a + b + c = 0

ab + bc + ca = 3

abc = -3

Since, a, b, c are the roots of equation,

a³ + 3a + 3 = 0

b³ + 3b + 3 = 0

c³ + 3c + 3 = 0

Thus, a³+ b³ + c³ + 3(a+b+c) + 9 = 0

implies a³ + b³ + c³ = -9(since a +b +c=0)

now,

(a+b+c)² = a² + b² + c² + 2(ab + bc+ ca)

0 = a² + b² + c² + 2(3)

Thus, a² + b² + c² = -6

here we can use the conditional identity

( a² + b² + c² )/2.( a³ + b³ + c³ )/3= a⁵ + b⁵ + c⁵/5

^{_{when a +b+c=0}}

now substituting values

we get -6/2.-9/3.5=a⁵ + b⁵ + c⁵

hence a⁵ + b⁵ + c⁵=-3.-3.5=45

hsbhatt

really beautiful shreyarya. I wish I could give you one more salute for this one.

shreyaarya

thank you sir

anchitsaini

shreyaarya
are u really below class tenth!!!!
amazing

shreyaarya

i'll be in tenth now(march 12)

thank you

hsbhatt

thats really impressive for any class

ashwin4u4ever

whooaa... shreyarya... thats really superb stuf..

karansingh

nice solution, shreyarya..............

karansingh

nice solution, shreyarya..............

in fact great

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