Find the value


Find the value

$x(x-2)(x-4)+4 x-8$




= x(x  - 2)(x - 4) + 4(x - 2)

Taking (x - 2) common in both the terms

=(x - 2){x(x - 4) + 4}

$=(x-2)\left\{x^{2}-4 x+4\right\}$

Now splitting the middle term of $x^{2}-4 x+4$

$=(x-2)\left\{x^{2}-2 x-2 x+4\right\}$






$\therefore x(x-2)(x-4)+4 x-8=(x-2)^{3}$



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