Question

plz do 4 and 5 help me

Answer 1

Hey !!!

Q . 4

solution:-

a + b + c = 9
ab + bc + ca = 26

•°• (a + b + c ) ² = a² + b² + c² + 2ab + 2bc + 2ac

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

9² - 2 ( 26) = a² + b² + c²

81 - 52 = a²+ b² + c²

29 = a² + b² + c²

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Q. 5

solution :-

if a + b + c = 0
then, we know that , a³ + b³ + c³ = 3abc

Now from Question

a² /bc+ b²/ca + c² /ab

a³ + b³ + c³ /abc ✔[lcm taken ]

3abc /abc = 3 Prooved

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Hope it helps you !!!

@Rajukumar111

Answer 2

qn 4.

a+b+c=9 and ab+bc+ca=26
(a+b+c)^2= a^2+b^2+c^2+2 (ab+bc+ca)
9^2 = (a^2+b^2+c^2)+2 (26)
so, a^2+b^2+c^2=(29)^1/2..

qn 5.

since a+b+c=0
so b+c= -a
now
(a+b+c)^3= a^3+(b+c)^3+3a (b+c)(a+b+c)
0 = a^3+b^3+c^3+3bc (b+c)
a^3+b^3+c^3= -3bc (b+c)
a^3+b^3+c^3 = 3abc.......... (1)

now come to the eq
a^2/bc+b^2/ca+c^2/ab=3

now eq (1) can be written as

(a^2×a)+(b^2×b)+(c^2×c)=3abc
(a^2×a+b^2×b+c^2×c)/abc = 3
(a^2×a)/abc + (b^2×b)/abc + (c^2×c)/abc=3

now cancel out the same in numerator and denominator

we get

a^2/bc + b^2/ac + c^2/ab = 3

hence proved....
hope it will help you.