Question

Evaluate x- a^3 + x- b^3+ x- c^3- 3 x -a x- b x- c given 3x = a + b +c

Answer 1

Answer:

0

Step-by-step explanation:

given equation is in form

p³+q³+r³ -3pqr = (p+q+r)(p²+q²+r²-pq-qr-rp)

p= x-a

q=x-b

r= x-c

p+q+r = 3x-(a+b+c)=3x-3x =0

p³+q³+r³ -3pqr = 0× anything = 0

Answer 2

SOLUTION

$\sf{3x = a + b + c}$

TO EVALUATE

$\sf{ {(x - a)}^{3} + {(x - b)}^{3} + {(x - c)}^{3} - 3(x - a)(x - b)(x - c)}$

FORMULA TO BE IMPLEMENTED

We are aware of formula :

If a + b + c = 0 then

$\sf{ {a}^{3} + {b}^{3} + {c}^{3} = 3abc}$

EVALUATION

$\sf{Let \: \: \: u= x - a \: ,\: v = x - b \: ,\: w = x - c}$

Then

$\sf{u + v + w}$

$\sf{ = (x - a) + (x - b) + (x - c)}$

$\sf{ = 3x - ( a + b + c)}$

$\sf{ = 0 \: \: \: \: \:( \: \because \: 3x = a + b + c \: )}$

$\therefore \: \sf{u + v + w = 0}$

So by the above mentioned formula

$\sf{ {u}^{3} + {v}^{3} + {w}^{ 3} = 3uvw }$

$\implies \: \sf{ {u}^{3} + {v}^{3} + {w}^{ 3} - 3uvw = 0 }$

$\sf{ \implies {(x - a)}^{3} + {(x - b)}^{3} + {(x - c)}^{3} - 3(x - a)(x - b)(x - c)} = 0$

FINAL ANSWER

$\sf{ {(x - a)}^{3} + {(x - b)}^{3} + {(x - c)}^{3} - 3(x - a)(x - b)(x - c)} = 0$

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