Add the following: a(a+b), b(b+c), c(a+c) and $a^{2}$ + $b^{2}$ + $c^{2}$

+ ab +

+ bc +

+ ac

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ ab +

+ bc +

+ ac

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 2ab +

+ bc +

+ ac

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 4ab +

+ 4bc +

+ 4ac

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A + ab + + bc + + ac
a(a+b) = $a^{2}$ + ab
b(b+c) = $b^{2}$ + bc
c(a+c) = ac + $c^{2}$
Now, $a^{2}$ + ab
+ $b^{2}$ + bc
+ $c^{2}$ + ac
+ $a^{2}$ + $b^{2}$ + $c^{2}$
= $2 a^{2}$ + ab + $2 b^{2}$ + bc + $2 c^{2}$ + ac

Suggest Corrections

Similar questions

Add the following: a(a+b), b(b+c), c(a+c) and $a^{2}$ + $b^{2}$ + $c^{2}$

Add the following:

$a (a + b), b (b + c), c (a + c) a n d a^{2} + b^{2} + c^{2}$ [2 MARKS]

\frac{a {(b - c)}^{2}}{(c - a) (a - b)} + \frac{b {(c - a)}^{2}}{(a - b) (b - c)} + \frac{c {(a - b)}^{2}}{(b - c) (c - a)} = a + b + c

∣ ∣ ∣ ∣ \begin{matrix} b + c & c + a & a + b c + a & a + b & b + c a + b & b + c & c + a \end{matrix} ∣ ∣ ∣ ∣ == 2 (a + b + c) (a b + b c + c a - a^{2} - b^{2} - c^{2})

Join BYJU'S Learning Program