Question

சுரு‌க்கு‌க
(2a+3b+4c)( (4a^2+9b^2+16c^2-┤6ab-12bc-8ca)

Answer 1

||✪✪ CORRECT QUESTION ✪✪||

prove That,

(2a+3b+4c)² =(4a^2+9b^2+16c^2+12ab+24bc+16ca)

|| ✰✰ ANSWER ✰✰ ||

Taking LHS,

→ (2a + 3b + 4c)²

Using (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca , now , we get,

→ (2a)² + (3b)² + (4c)² + 2*2a*3b + 2*3b*4c + 2*4c*2a

→ 4a² + 9b² + 16c² + 12ab + 24bc + 16ca = RHS.

✪✪ Hence Proved ✪✪

Answer 2

விளக்கம்:

$(2a+3b+4c) (4a^2+9b^2+16c^2-6ab-12bc-8ca)$

$&(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c\right. - ca)$

$=a^{3}+b^{3}+c^{3}-3 a b c$

$\begin{aligned}&a=2 a, b=3 b, c=4 c\\&(2 a+3 b+4 c)\left(4 a^{2}+9 b^{2}+16 c^{2}-6 a b-12 b c-8 c a)\end{aligned}$

$=a^{3}+b^{3}+c^{3}-3 a b c$

$a=x, b=-2 y, c=3 z$

$\begin{aligned}&(x-2 y+3 z)\left(x^{2}+4 y^{2}+9 z^{2}+2 x y+6 y z-3 x z)\end{aligned}$

$\begin{aligned}&=x^{3}+(-2 y)^{3}+(3 z)^{3}-3(x)(-2 y)(3 z)\end{aligned}$

$=x^{3}-8 y^{3}+27 z^{3}+18 x y$