Question

If the area of the rectangle is (a3 + b3 + c3 - 3abc) sq units and the length of the rectangle is (a+b+c) units. Find its breadth.
how is it coming (a²+b²+c²-ab-bc-ca)?​

Answer 1

By rearranging and grouping the terms we get,

⇒ (a² – 2ab + b²) + (b² – 2bc + c²) + (c² – 2ca + a²) = 0

⇒ (a – b)² + (b – c)² + (c – a)² = 0 (Since, (a – b)² = (a² – 2ab + b²))

As the sum of all the three squares is zero thus, each term will be equal to zero

Therefore,

(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

Thus, a = b = c.

Hence,

(c + a)/b = (b+b)/b

= 2b/b

= 2

So, If a²+b²+c² = ab+bc+ca, (c+a)/b = 2.