Question

plzz help me
If it is given that
(a+b)²+ (b+c)² + (c+d) ² =
4 (ab + bc + cd) answer is a=b=c​

Answer 1

Correct Question

If it is given that

(a+b)²+ (b+c)² + (c+d) ² + (a+d)² =

6(ab + bc + cd+ ad) answer is a=b=c

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• (a+b)²+ (b+c)² + (c+d) ² =6(ab + bc + cd+ad)

${\sf{\green{\underline{\large{To\:Prove}}}}}$

• a=b=c

${\sf{\pink{\underline{\Large{Explanation}}}}}$

(a+b)²+ (b+c)² + (c+d) ² =6(ab + bc + cd)

We know that

=>(a+b)² = a² + b² +2ab

=>(c+b)² = b² + c² +2cb

=>(c+d)² = c² + d² +2cd

=>(a+d)² = a²+d² +2ad

Adding all

(a+b)²+ (b+c)² + (c+d) ²+(a+d)² =a² + b² +2ab+b² + c² +2cb+c² + d² +2cd

=>a² + b² +2ab+b² + c² +2cb+c² + d² +2cd+a²+d²+2ad=6(ab + bc + cd)

=>a² + b² +2ab+b² + c² +2cb+c² + d² +2cd+a²+d²+2ad=6ab + 6bc + 6cd +6ad

=>a² + b² +2ab+b² + c² +2cb+c² + d² +2cd+a²+d²+2ad-(6ab + 6bc + 6cd +6ad)=0

=>a² + b² +2ab+b² + c² +2cb+c² + d² +2cd+a²+d²+2ad-6ab - 6bc - 6cd - 6ad=0

=>a² + b²+b² + c² +c² + d² +a²+d²+-4ab - 4bc - 4cd - 4ad=0

=>(a-b)²+ (b-c)² + (c-d) ²+(a-d)² =0

Taking

(a-b)² = 0

=>a=b

Similarly

b=c , c=d and a =d

From this

a=b=c=d

Answer 2

Step-by-step explanation:

(a+b)²+ (b+c)² + (c+d) ² =6(ab + bc + cd)

We know that

=>(a+b)² = a² + b² +2ab

=>(c+b)² = b² + c² +2cb

=>(c+d)² = c² + d² +2cd

=>(a+d)² = a²+d² +2ad

Adding all

(a+b)²+ (b+c)² + (c+d) ²+(a+d)² =a² + b² +2ab+b² + c² +2cb+c² + d² +2cd

=>a² + b² +2ab+b² + c² +2cb+c² + d² +2cd+a²+d²+2ad=6(ab + bc + cd)

=>a² + b² +2ab+b² + c² +2cb+c² + d² +2cd+a²+d²+2ad=6ab + 6bc + 6cd +6ad

=>a² + b² +2ab+b² + c² +2cb+c² + d² +2cd+a²+d²+2ad-(6ab + 6bc + 6cd +6ad)=0

=>a² + b² +2ab+b² + c² +2cb+c² + d² +2cd+a²+d²+2ad-6ab - 6bc - 6cd - 6ad=0

=>a² + b²+b² + c² +c² + d² +a²+d²+-4ab - 4bc - 4cd - 4ad=0

=>(a-b)²+ (b-c)² + (c-d) ²+(a-d)² =0

Taking

(a-b)² = 0

=>a=b

Similarly

b=c , c=d and a =d

From this

a=b=c=d

$\huge \: thx$