Concept:

Let two vectors are \(\rm \vec {a}\) and \(\rm \vec {b}\)

Magnitude of sum of  \(\rm \vec {a}\) and \(\rm \vec {b}\):

\(\rm \left|\vec a+\vec b\right| = \sqrt{a^2+b^2+2ab\cosθ}\)

Magnitude of difference of  \(\rm \vec {a}\) and \(\rm \vec {b}\):

\(\rm \left|\vec a-\vec b\right| = \sqrt{a^2+b^2-2ab\cosθ}\)

where a, b are magnitude of vectors \(\rm \vec a \text{ and } \vec b\); and θ is angle between them.

Calculation:

Given:

\(\rm \left|\vec a\right|=\alpha\text{, }\left|\vec b\right|=\alpha\text{ and }\left|\vec a+\vec b\right|=\alpha\)

As we know,

\(\rm \left|\vec a+\vec b\right| = \sqrt{a^2+b^2+2ab\cosθ}\)

⇒ \(\rm \alpha = \sqrt{\alpha^2+\alpha^2+2(\alpha)(\alpha)\cosθ}\)

⇒ \(\rm \alpha^2 = 2\alpha^2+2\alpha^2\cosθ\)

⇒ \(\rm -1=2\cosθ\)

⇒ \(\boldsymbol{\rm \cosθ=-\frac{1}{2}}\)

Now, 

\(\rm \left|\vec a-\vec b\right| = \sqrt{a^2+b^2-2ab\cosθ}\)

⇒ \(\rm \left|\vec a-\vec b\right| = \sqrt{\alpha^2+\alpha^2-2(\alpha)(\alpha)\cosθ}\)

\(∵ \cos θ = -\frac{1}{2}\)

⇒ \(\rm \left|\vec a-\vec b\right| = \sqrt{2\alpha^2-2\alpha^2 (\frac{-1}{2})}\)

⇒ \(\rm \left|\vec a-\vec b\right| = \sqrt{2\alpha^2+\alpha^2}\)

⇒ \(\boldsymbol{\rm \left|\vec a-\vec b\right| = \sqrt{3}\alpha}\)