Question

Evaluate :- [7{(8)^ ⅓ +(125) ^⅓}^2]⅓
Answer it...its urgent...please ​

Answer 1

Step-by-step explanation:

Evaluate :- [7{(8)^ ⅓ +(125) ^⅓}^2]⅓

Answer it...its urgent...please

Answer 2

Answer:

Question :

Evaluate :-

${\implies{\Bigg[ 7\bigg\{{(8)}^{\frac{1}{3}} + {(125)}^{\frac{1}{3} }\bigg\}^{2} \Bigg]^{\frac{1}{3}}}}$

Solution :

${ = {\Bigg[ 7\bigg\{{(8)}^{\frac{1}{3}} + {(125)}^{\frac{1}{3} }\bigg\}^{2} \Bigg]^{\frac{1}{3}}}}$

${ = {\Bigg[ 7\bigg\{{({2}^{3})}^{\frac{1}{3}} + {({5}^{3})}^{\frac{1}{3} }\bigg\}^{2} \Bigg]^{\frac{1}{3}}}}$

Using law of exponent rule to continuously evaluating the equation : (aᵐ)ⁿ = aᵐⁿ

${ = {\Bigg[ 7\bigg\{{({2})}^{3 \times \frac{1}{3}} + {({5})}^{3 \times \frac{1}{3} }\bigg\}^{2} \Bigg]^{\frac{1}{3}}}}$

${ = {\Bigg[ 7\bigg\{{({2})}^{\cancel{3} \times \frac{1}{\cancel{3}}} + {({5})}^{\cancel{3} \times \frac{1}{\cancel{3}}}\bigg\}^{2} \Bigg]^{\frac{1}{3}}}}$

${ = {\Bigg[ 7\bigg\{2 + 5\bigg\}^{2} \Bigg]^{\frac{1}{3}}}}$

${ = {\Bigg[ 7\bigg\{ \: \: 7 \: \: \bigg\}^{2} \Bigg]^{\frac{1}{3}}}}$

${ = {\Bigg[ 7\bigg\{ \: 7 \times 7 \: \bigg\} \Bigg]^{\frac{1}{3}}}}$

${ = {\Bigg[ 7 \times 7 \times 7\Bigg]^{\frac{1}{3}}}}$

${ = {\Big[ \: \: {7}^{3} \: \: \Big]^{\frac{1}{3}}}}$

Again using law of exponent rule to continuously evaluating the equation : (aᵐ)ⁿ = aᵐⁿ

${ = {\Big[ \: \: {7} \: \: \Big]^{3 \times \frac{1}{3}}}}$

${ = {\Big[ \: \: {7} \: \: \Big]^{\cancel{3} \times \frac{1}{\cancel{3}}}}}$

${ = {\Big[ \: \: {7} \: \: \Big]^{1}}}$

${={\sf{\underline{\underline{\red{Ans = 7 }}}}}}$

Hence, the answer is 7.

$\rule{220pt}{2.5pt}$

Learn More :

☼ Algebraic identities:-

• ➛ (a+b)²+(a-b)² = 2a²+2b²
• ➛ (a+b)²-(a-b)² = 4ab
• ➛ (a+b)(a -b) = a²-b²
• ➛ (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
• ➛ (a-b)³ = a³-b³-3ab(a-b)
• ➛ (a³+b³) = (a+b)(a²-ab+b²)
• ➛ a²+b² = (a+b)²-2ab
• ➛ a³-b³ = (a-b)(a²+ab +b²)
• ➛ If a + b + c = 0 then a³ + b³ + c³ = 3abc

☼ BODMAS :

↝ BODMAS rule is an acronym used to remember the order of operations to be followed while solving expressions in mathematics.

It stands for :-

• »» B - Brackets,
• »» O - Order of powers or roots,
• »» D - Division,
• »» M - Multiplication 
• »» A - Addition
• »» S - Subtraction.

↝ It means that expressions having multiple operators need to be simplified from left to right in this order only.

☼ BODMAS RULE :

↝ First, we solve brackets, then powers or roots, then division or multiplication (whatever comes first from the left side of the expression), and then at last subtraction or addition.

• ↠ Addition (+)
• ↠ Subtraction (-)
• ↠ Multiplication (×)
• ↠ Division (÷)
• ↠ Brackets ( )

☼ EXPONENT :

↝ The exponent of a number says how many times to use the number in a multiplication.

☼ LAW OF EXPONENT :

The important laws of exponents are given below:

• ➠ ${\rm{{a}^{m} \times {a}^{n} = {a}^{m + n}}}$

• ➠ ${\rm{{a}^{m}/{a}^{n} = {a}^{m - n}}}$

• ➠ ${\rm{({a}^{m})^{n} = {a}^{mn}}}$

• ➠ ${\rm{{a}^{n}/{b}^{n} = ({a/b})^{n} }}$

• ➠ ${\rm{{a}^{0} = 1}}$

• ➠ ${\rm{{a}^{ - m} = {1/a}^{m}}}$

• ➠ ${\rm{{a}^{\frac{1}{n} } = \sqrt[n]{a}}}$

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