Question

If a+b-c= 10 then prove that 2a+2b-2c=✓10​

Answer 1

Step-by-step explanation:

Consider, a

2

+b

2

+c

2

–ab–bc–ca=0

Multiply both sides with 2, we get

2(a

2

+b

2

+c

2

–ab–bc–ca)=0

⇒ 2a

2

+2b

2

+2c

2

–2ab–2bc–2ca=0

⇒ (a

2

–2ab+b

2

)+(b

2

–2bc+c

2

)+(c

2

–2ca+a

2

)=0

⇒ (a–b)

2

+(b–c)

2

+(c–a)

2

=0

Since the sum of square is zero then each term should be zero

⇒ (a–b)

2

=0,(b–c)

2

=0,(c–a)

2

=0

⇒ (a–b)=0,(b–c)=0,(c–a)=0

⇒ a=b,b=c,c=a

∴ a=b=c.

Answer 2

Corrected Question:-

If a+b-c = 10 then prove that 2a+2b-2c = 20

Given :-

♦ a + b - c = 10

Required To Prove :-

♦ 2a + 2b - 2c = 20

Solution :-

Given that

a+b-c = 10

On multiplying with 2 both sides then

=> 2(a+b-c) = 2×10

=> 2a+2b-2c = 20

Hence, Proved.

Additional Information :-

♦ (a+b)² = a²+2ab+b²

♦ (a-b)² = a²-2ab+b²

♦ (a+b)(a-b) = a²-b²

♦ (a+b+c)² = a²+b²+c²+2ab+2bc+2ca

♦ (a+b)³ = a³+b³+3a²b+3ab²

♦ (a-b)³ = a³-b³-3a²b+3ab²