Question

I don't know how to solve...Can anyone help me??​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\: {x}^{3} {y}^{2}z \: = \: 2160$

Let's find the prime factorization of 2160.

So,

Prime factorization of 2160

$\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{5}}}&{\underline{\sf{\:\:2160 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:432 \:\:}} \\\underline{\sf{2}}&\underline{\sf{\:\:216\:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:108 \:\:}} \\ {\underline{\sf{2}}}& \underline{\sf{\:\:54 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:27 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:9 \:\:}} \\ {\underline{\sf{3}}}& \underline{\sf{\:\:3 \:\:}} \\\underline{\sf{}}&{\sf{\:\:1 \:\:}} \end{array}\end{gathered}$

$\rm :\longmapsto\:2160 = {3}^{3} \times {2}^{4} \times 5$

can be rewritten as

$\rm :\longmapsto\:2160 = {3}^{3} \times {4}^{2} \times 5$

So,

$\rm :\longmapsto\: {x}^{3} {y}^{2}z \: = \: {3}^{3} \times {4}^{2} \times 5$

So, on comparing we get

$\rm :\longmapsto\:x = 3$

$\rm :\longmapsto\:y = 4$

$\rm :\longmapsto\:z = 5$

Hence,

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + {z}^{2}$

$\rm \:  =  \: {3}^{2} + {4}^{2} + {5}^{2}$

$\rm \:  =  \:9 + 16 + 25$

$\rm \:  =  \:50$

More Identities to know :-

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)

Answer 2

Given :-

x³y²z = 2160

To Find :-

x² + y² + z²

Solution :-

x³y²z = 2160

Prime factorizing 2160

2 | 2160

2 | 1080

2 | 540

2 | 270

3 | 135

3 | 45

3 | 15

3 | 5

5 | 1

2160 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5

2160 = 4¹ × 4¹ × 3³ × 5

2160 = 4⁽¹ ⁺ ¹⁾ × 3³ × 5

2160 = 4² × 3³ × 5

Now

x³y²z = 4² × 3³ × 5

x³ = 3³

x = 3

y² = 4²

y = 4

z = 5

Now

x² + y² + z²

(3)² + (4)² + (5)²

9 + 16 + 25

9 + 41

50