Question

If a = 5 ,b = 6 and c = 10 , find the value of: a+b_c-a​

Answer 1

(a+b)=6k

(b+c)=7k

(c+a)=8k

Now, we get addition of LHS & RHS

2a+2b+2c=21k

2(a+b+c)=21k

2(14)=21k —————- [given - (a+b+c)=14]

28=21k

k=28/21=4/3 ———————(1)

Since,

a+b+c=14

c=14-(a+b)

we have,

a+b=6k

a+b=6(4/3) ——————-from 1

a+b=8

now,

c=14-(a+b)

c=14–8

c=6

Step-by-step explanation:

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Answer 2

We have,

a+b+c=5

ab+bc+ca=10

Since,

(a+b+c)

2

=25

a

2

+b

2

+c

2

+2(ab+bc+ca)=25

a

2

+b

2

+c

2

+2×10=25

a

2

+b

2

+c

2

=5

We know that,

a

3

+b

3

+c

3

−3abc=(a+b+c)(a

2

+b

2

+c

2

−(ab+bc+ca))

a

3

+b

3

+c

3

−3abc=(5)×(5−10)

a

3

+b

3

+c

3

−3abc=5×−5

a

3

+b

3

+c

3

−3abc=−25

Hence, proved.