For terms $a^{2} b^{2}, a^{3} b^{3}$ and $a^{3} b^{2}$ , $a^{3} b^{3}$ is known as ___

Lowest common multiple

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

Highest common multiple

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

Highest common factor

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

Lower common factor

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

Open in App

Solution

The correct option is A
Lowest common multiple

$a^{2} b^{2}, a^{3} b^{3}$ and $a^{3} b^{2}$ are common multiples of $a^{2}$ and $b^{2}$ and here $a^{2} b^{2}$ is of the lowest power. So $a^{2} b^{2}$ is the lowest common multiple.

Suggest Corrections

Similar questions

a^{2} b^{2} - a^{3} b^{3} + 2 a b

| a | < 1, | b | < 1

1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} +

| a | < 1

| b | < 1

s = 1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + . . . i s - .

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Δ

1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + . . . . b

Join BYJU'S Learning Program