# Thread: d2 = a2 + b2 + c2

1. ## d2 = a2 + b2 + c2

If d^2 = a^2 + b^2 + c^2 , then show that (a + b + c + 2d)^2 = (a + b + d)^2 + (a + c + d)^2 + (b + c + d)^2

2. Originally Posted by pacman
If d^2 = a^2 + b^2 + c^2 , then show that (a + b + c + 2d)^2 = (a + b + d)^2 + (a + c + d)^2 + (b + c + d)^2
HI

Let $\displaystyle a+b+c+d=k$ ---1

$\displaystyle (a + b + d)^2 + (a + c + d)^2 + (b + c + d)^2$

$\displaystyle =(k-c)^2+(k-b)^2+(k-a)^2$

$\displaystyle =k^2-2kc+c^2+b^2-2kb+b^2+k^2-2ka+a^2$

$\displaystyle =3k^2-2k(k-d)+d^2$

$\displaystyle =k^2+2kd+d^2$

$\displaystyle =(k+d)^2$

then from 1

$\displaystyle =(a+b+c+d+d)^2$

$\displaystyle =(a+b+c+2d)^2$

Hence shown

3. mathaddict, i like your substitution method, by letting . Better than my manual expansion of the squared terms and cancelations . . . .

4. Originally Posted by pacman
mathaddict, i like your substitution method, by letting . Better than my manual expansion of the squared terms and cancelations . . . .
yeah , the working can be very tedious if you were to expand the whole thing .

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