Question

6ab2c3× 4b-2c-3d is =​

Answer 1

Correct question is "6ab²c³×( 4b-2c-3d) is = ?"

Answer:

Required answer is $24a {b}^{3} {c}^{3} - 12a {b}^{2} {c}^{4} - 18a{b}^{2} {c}^{3}d$

Step-by-step explanation:

Given,

$6 a{b}^{2} {c}^{3} \times (4b - 2c - 3d)$

Now by Algebra's multiplication,

$6a {b}^{2} {c}^{3} \times 4b - 6a {b}^{2} {c}^{3} \times 2c - 6a {b}^{2} {c}^{3} \times 3d$

$= 24a {b}^{2 + 1} {c}^{3} - 12a {b}^{2} {c}^{3 + 1} - 18a{b}^{2} {c}^{3}d$

$= 24a {b}^{3} {c}^{3} - 12a {b}^{2} {c}^{4} - 18a{b}^{2} {c}^{3}d$

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

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Answer 2

$\displaystyle \sf 6a {b}^{2} {c}^{3} \times 4 {b}^{ - 2} {c}^{ - 3} d = \bf \:24ad$

Correct question :

$\displaystyle \sf 6a {b}^{2} {c}^{3} \times 4 {b}^{ - 2} {c}^{ - 3} d = ?$

Given :

$\displaystyle \sf 6a {b}^{2} {c}^{3} \times 4 {b}^{ - 2} {c}^{ - 3} d$

To find :

To simplify the expression

Formula Used :

We are aware of the formula on indices that :

$\sf{1. \: \: {a}^{m} \times {a}^{n} = {a}^{m + n} }$

$\displaystyle \sf{2. \: \: {a}^{0} = 1}$

Solution :

Step 1 of 2 :

Write down the given expression

Here the given expression is

$\displaystyle \sf 6a {b}^{2} {c}^{3} \times 4 {b}^{ - 2} {c}^{ - 3} d$

Step 2 of 2 :

Simplify the given expression

$\displaystyle \sf 6a {b}^{2} {c}^{3} \times 4 {b}^{ - 2} {c}^{ - 3} d$

$\displaystyle \sf = (6 \times 4) \times a \times ( {b}^{2} \times {b}^{ - 2} ) \times ( {c}^{3} \times {c}^{ - 3} ) \times d$

$\displaystyle \sf = 24 \times a \times {b}^{(2 - 2)} \times {c}^{(3 - 3)} \times d\: \: \: \bigg[ \: \because \:{a}^{m} \times {a}^{n} = {a}^{m + n} \bigg]$

$\displaystyle \sf = 24 \times a \times {b}^{0} \times {c}^{0} \times d$

$\displaystyle \sf = 24 \times a \times 1 \times 1 \times d\: \: \: \bigg[ \: \because \: {a}^{0} = 1\bigg]$

$\displaystyle \sf = 24ad$

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