Question

Question for Maths Students...

Prove that
$\orange{ \large{ \dag}} \red{ \tt \: (x + a)(x + b)(x + c) = {x}^{3} + {x}^{2} (a + b + c) + x(ab + bc + ac) + abc}$
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Answer 1

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

$\sf (x + a)(x + b)(x + c)$

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Prove:-}}}}}}}$

$\sf (x + a)(x + b)(x + c) = x^3 + x^2 (a + b + c) + x (ab + bc + ac) + abc$

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Proof:-}}}}}}}$

$\sf (x + a)(x + b)(x + c)$

• First let's take (x + a)(x + b),

$\sf (x + a)(x + b)$

$\sf x (x + b) + a (x + b)$

$\sf x^2 + bx + ax + ab$

• Now, let's multiply the above one with $\sf (x + c)$

$\sf (x² + bx + ax + ab)(x + c)$

$\sf x (x^2 + bx + ax + ab) + c (x^2 + bx + ax + ab)$

$\sf x^3 + bx^2 + ax^2 + abx + cx^2 + bcx + acx + abc$

• Rearranging the terms,

$\sf x^3 + ax^2 + bx^2 + cx^2 + abx + bcx + acx + abc$

• Taking out common ones i.e x² and x,

$\sf x^3 + \underline{ax^2 + bx^2 + cx^2} + \underline{abx + bcx + acx} + abc$

$\sf x^3 + x^2 (a + b + c) + x (ab + bc + ac) + abc$

• Therefore,

$\underline{\boxed{\purple{\tt (x + a)(x + b)(x + c) = x^{3} + x^{2} (a + b + c) + x(ab + bc + ac) + abc}}}$

Hence, proved.

Answer 2

Solution!!

(x + a)(x + b)(x + c) = x³ + x²(a + b + c) + x(ab + bc + ac) + abc

Taking LHS,

= (x + a)(x + b)(x + c)

Multiplying the parentheses

= (x² + bx + ax + ab)(x + c)

Multiply the parentheses

= x³ + cx² + bx² + bcx + ax² + acx + abx + abc

Reordering the terms

= x³ + ax² + bx² + cx² + abx + bcx + acx + abc

Taking x² as common in 2nd, 3rd and 4th term

= x³ + x²(a + b + c) + abx + bcx + acx + abc

Taking x as common in 3rd, 4th and 5th term

= x³ + x²(a + b + c) + x(ab + bc + ac) + abc

LHS = RHS

Hence, proved.