Question

If a + b + c = 5 and ab + bc + ca = 10, then prove that a

3

+ b

3

+ c

3

–3abc = – 25.​

Answer 1

Step-by-step explanation:



1

Secondary School 

 

Math 

 

5 points

If a+b+c=5 and ab+bc+ca=10 then prove that a3+b3+c3-3abc= -25 .

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 by Tushar241 22.05.2017

Answers



abhi178 

 

Genius

We know ,

a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca)

now ,

a + b + c = 5

ab + bc + ca = 10

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

(5)² -2×10 = a² + b² + c²

a² + b² + c² =5

hence ,

a³ + b³ +c³ -3abc = ( a + b + c )(a² + b² + c² -ab- bc-ca)

=( 5)( 5 - 10) = 5 × (-5) = -25

Answer 2

The sum of ${\displaystyle \sf \sum \limits^{100}_ {k\ =\ 1}\ \dfrac{k}{k^4\ +\ k^2\ +\ 1}}$ is equal to :-

$\sf {(A)\ \dfrac{4950}{10101}}$

$\sf {(B)\ \dfrac{5050}{10101}}$

$\sf {(C)\ \dfrac{5151}{10101}}$

$\sf {456\ +\ \dots \dots \dots}$