pls answer this question no spam will be reported
Answers
Answer:
→ Hey Mate ,
→ Given Question : If a² + b + c = 1
→ : a² + b² + c² = 2
→ : a³ + b³ + c³ = 3
→ Then a ^ 5 + b^5 + c ^5 = ?
Step-by-step explanation:
→ Solution : ( a² + b² + c² ) ( a³ + b³ + c³ ) ( observation )
→ : + α² b³ + α ² c³
→ : α³b² + + b² c³
→ : α²c³ + b³ c²+ c^5
→ : α^5 + b^5 + c^5 + α²b² ( α+b) + b² + c²( b +c )+α²c²(α + c)
→ : α^5 + b^5 + c^5 + α²b² +b²c²+ α²c²- α²b² - αb²c² - α²bc²
→ : αbc( αb + bc + αc
→ ( observation) : ( α + b + c )² = ( 1 ) ²
→ : α² + αb +ac + ab + b² + bc + ac + bc + c²
→ : α² + b² + c² + Z ( ab + bc + ac) = 1
→ : ( α² + b² + c² ) 2 + 2 ( ab + bc + ac)= 1
→ Know :[ αb + bc + ac = - 1 / 2 ]
→ observation : ( α + b + c) ³ = ( 1 /3 ) ³ [ ∵ trinomial expansion ]
→ : here used trinomial expansion refer attachment
→ observation : ( α + b +c ) ³ = ( 1 ) ³
→ : α ^5 + 3a²b + 3 ab² + b³ + 3a ²c + babc + 3b²c → +3ac²+c³ = 1
→ α³ + b³ + c³ + 3 ( α² ( b + c ) + b²( a +c )+ c² (a +b)
→ 6abc + 3 + 3 ( α² +b²+c² )
--------------------- ( this sign is of division )
→ ( - α³ -b³-c³ )
→ = 1
→ 6abc +3 ( 2 - (3) ) = -2
→ 6abc -3 = -2
→ Know [ αbc = 1/6 ]
→ a^5 + b^5 + c^5 = ( a² + b² + c² ) ( a ³ + b³ + c³) +abc (ab + bc + ca )
→ 2 .3 + 1 . ( -1 )
→ ------------
→ 6 . 2
→ observation : (αb + bc + ac) ² = (-1 /2 ) ( αb + bc + ac ) (αb + bc + bc)
→ α²b² + αb²c +a²bc
→ αb²c+ b² c² + a bc² = 1 / 4
→ α² b²c + abc² +α²c²
→ α²b² + b²c² +α ² c² +2(abc) (b + a +c ) 1/4
→ α²b² + b²c² +α ² c² (1 / 6 ) ( 1 ) = 1/4
→ Final Answer = > - 1 / 2 finally done.
---------------------------------------------------------------------------------------------------
Given : a + b + c = 1
a² + b² + c² = 2
a³ + b³ + c³ = 3
To Find : a⁵ + b⁵ + c⁵
Solution:
a² + b² + c² = 2
a³ + b³ + c³ = 3
Multiplying both
a⁵ + b⁵ + c⁵ + a²( b³ + c³) + b²(a³ + c³) + c²(a³ + b³ ) = 6
=> a⁵ + b⁵ + c⁵ + a²b² (a + b) +a²c²(a + c) + b²c²(b + c) = 6
=> a⁵ + b⁵ + c⁵ + a²b² (1-c) +a²c²(1-b) + b²c²(1-a) = 6
=> a⁵ + b⁵ + c⁵ + a²b² +a²c² + b²c² - abc(ab + ac + bc) = 6 Eq1
a + b + c = 1
squaring both sides
=> a² + b² + c² + 2(ab + ac + bc) = 1
=> 2 + 2(ab + ac + bc) = 1
=> ab + ac + bc = - 1/2
Squaring both sides
=> a²b² +a²c² + b²c² + 2abc(a + b + c) = 1/4
=> a²b² +a²c² + b²c² + 2abc = 1/4
=> a²b² +a²c² + b²c² = 1/4 - 2abc
a⁵ + b⁵ + c⁵ + a²b² +a²c² + b²c² - abc(ab + ac + bc) = 6
=> a⁵ + b⁵ + c⁵ + 1/4 - 2abc - abc (-1/2) = 6
=> a⁵ + b⁵ + c⁵ = 23/4 + 3abc/2
a + b + c = 1
a² + b² + c² = 2
Multiplying both
a³ + b³ + c³ + a(b² + c² ) + b(a² + c²) + c(a² + b²) = 2
=> 3 + ab (a + b) + ac(a + c) + bc(b + c) = 2
=> ab(1 - c) + ac( 1 - b) + bc( 1 - a) = - 1
=> ab + ac + bc -3abc = - 1
ab + ac + bc = - 1/2
=> - 1/2 - 3abc = - 1
=> 3abc = 1/2
=> abc = 1/6
a⁵ + b⁵ + c⁵ = 23/4 + 3abc/2
=> a⁵ + b⁵ + c⁵ = 23/4 + 1/4
=> a⁵ + b⁵ + c⁵ = 6
Learn More :
a+b+c=3 a^2+b^2+c^2=6 1/a+1/b+1/c=1 find values of a , b and c ...
https://brainly.in/question/8061154