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Q.$\text{If F(x)=}\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)},$
Show that F(0)=1.

Dear student,
given $F\left(x\right)=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}$
as we can see $F\left(a\right)=\frac{\left(a-b\right)\left(a-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(a-c\right)\left(a-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(a-a\right)\left(a-b\right)}{\left(c-a\right)\left(c-b\right)}=1$  ............(1)
also $F\left(b\right)=\frac{\left(b-b\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(b-c\right)\left(b-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(b-a\right)\left(b-b\right)}{\left(c-a\right)\left(c-b\right)}=1$......................(2)
also $F\left(c\right)=\frac{\left(c-b\right)\left(c-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(c-c\right)\left(c-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(c-a\right)\left(c-b\right)}{\left(c-a\right)\left(c-b\right)}=1$..........(3)
as we can see F(x) is a quadratic equation.and F(x)-1=0 is also a quadratic equation
as we can see from (1),(2),(3) x=a,b,c satisfy the equation.F(x)-1=0.
but quadratic equation has exactly two roots
here more than two values of x satisfies the equation.
so F(x)-1=0 is an identity.
$\Rightarrow$ it is satisfied by all values of x
$\Rightarrow$F(0)-1=0 is also true
$\Rightarrow$F(0)=1

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