CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Multiplication of Three or More Monomials

Multiply:vi 4...

Question

Multiply:
(vi) $4 a b c - 5 a^{2} b c - 6 a b^{2} c$ and $- 2 a b c^{2}$

Open in App

Solution

We have,
$(4 a b c - 5 a^{2} b c - 6 a b^{2} c) \times (- 2 a b c^{2})$

$= (4 a b c) \times (- 2 a b c^{2}) + (- 5 a^{2} b c) (- 2 a b c^{2}) + (- 6 a b^{2} c) (- 2 a b c^{2})$

$= - 8 a^{1 + 1} b^{1 + 1} c^{1 + 2} + 10 a^{2 + 1} b^{1 + 1} c^{1 + 2} + 12 a^{1 + 1} b^{2 + 1} c^{1 + 2}$

$= - 8 a^{2} b^{2} c^{3} + 10 a^{3} b^{2} c^{3} + 12 a^{2} b^{3} c^{3}$

$∴ (4 a b c - 5 a^{2} b c - 6 a b^{2} c) \times (- 2 a b c^{2}) = - 8 a^{2} b^{2} c^{3} + 10 a^{3} b^{2} c^{3} + 12 a^{2} b^{3} c^{3}$

Suggest Corrections

thumbs-up

4

Similar questions

Question 81 (i)

Add :

$7 a^{2} b c, - 3 a b c^{2}, 3 a^{2} b c, 2 a b c^{2}$

Show that:

$∣ ∣ ∣ ∣ \begin{matrix} b + c & a & a b & c + a & b c & c & a + b \end{matrix} ∣ ∣ ∣ ∣$ $= 4 a b c$

3 a^{2} b c + 6 a b^{2} c + 9 a b c^{2}

Q. Multiply:

vi)

2 \frac{1}{14}

- 7

Question 81 (i)

Add :

$7 a^{2} b c, - 3 a b c^{2}, 3 a^{2} b c, 2 a b c^{2}$

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Multiplication of Three or More Monomials

Standard VIII Mathematics

Join BYJU'S Learning Program

CrossIcon