Find the value of (1.6) 3 - (0.9) 3 - (0.7) 3.
3.024
Finding the Value of a Cubic Expression
The problem asks us to find the value of the expression $\left(1.6\right)^3 - \left(0.9\right)^3 - \left(0.7\right)^3$. We can solve this problem in two main ways: by calculating each cube directly and then subtracting, or by using an algebraic identity.
Method 1: Direct Calculation of Cubes
We will calculate the value of each term separately and then perform the subtraction.
- First, calculate $\left(1.6\right)^3$:
$\left(1.6\right)^3 = 1.6 \times 1.6 \times 1.6$ - $1.6 \times 1.6 = 2.56$
- $2.56 \times 1.6 = 4.096$
So, $\left(1.6\right)^3 = 4.096$.
- Next, calculate $\left(0.9\right)^3$:
$\left(0.9\right)^3 = 0.9 \times 0.9 \times 0.9$ - $0.9 \times 0.9 = 0.81$
- $0.81 \times 0.9 = 0.729$
So, $\left(0.9\right)^3 = 0.729$.
- Finally, calculate $\left(0.7\right)^3$:
$\left(0.7\right)^3 = 0.7 \times 0.7 \times 0.7$ - $0.7 \times 0.7 = 0.49$
- $0.49 \times 0.7 = 0.343$
So, $\left(0.7\right)^3 = 0.343$.
Now, substitute these values back into the original expression:
Value $= \left(1.6\right)^3 - \left(0.9\right)^3 - \left(0.7\right)^3$
Value $= 4.096 - 0.729 - 0.343$
Perform the subtractions:
Value $= 3.367 - 0.343$
Value $= 3.024$
Method 2: Using Algebraic Identity for Cubic Expressions
Observe the numbers in the expression: $1.6$, $0.9$, and $0.7$. Notice that $0.9 + 0.7 = 1.6$.
Let $a = 0.9$ and $b = 0.7$. Then $a+b = 0.9 + 0.7 = 1.6$.
The expression can be written as $\left(a+b\right)^3 - a^3 - b^3$.
We know the algebraic identity for the cube of a sum:
$\left(a+b\right)^3 = a^3 + b^3 + 3ab\left(a+b\right)$
Rearranging this identity, we get:
$\left(a+b\right)^3 - a^3 - b^3 = 3ab\left(a+b\right)$
Now, substitute the values of $a$ and $b$ back into the simplified expression $3ab\left(a+b\right)$:
$a = 0.9$
$b = 0.7$
$a+b = 1.6$
Value $= 3 \times \left(0.9\right) \times \left(0.7\right) \times \left(1.6\right)$
Calculate the product:
- $3 \times 0.9 = 2.7$
- $2.7 \times 0.7 = 1.89$
- $1.89 \times 1.6$
Let's perform the final multiplication:
| 1.89 | |
| × | 1.6 |
| --- | |
| 1134 | |
| + | 1890 |
| --- | |
| 3024 |
The total number of decimal places in $1.89$ (two) and $1.6$ (one) is three. So, the result should have three decimal places.
$1.89 \times 1.6 = 3.024$
Both methods give the same value.
Comparing Results with Options
The calculated value is $3.024$. Let's compare this with the given options:
- Option 1: $3.024$
- Option 2: $3.24$
- Option 3: $-3.24$
- Option 4: $-3.024$
The calculated value $3.024$ matches Option 1.
Conclusion on Finding the Value
The value of the expression $\left(1.6\right)^3 - \left(0.9\right)^3 - \left(0.7\right)^3$ is $3.024$. Using the algebraic identity $\left(a+b\right)^3 - a^3 - b^3 = 3ab\left(a+b\right)$ is generally more efficient and less error-prone than calculating the cubes directly, especially for competitive exams.
Revision Table: Key Concepts
| Concept | Description | Example (Relevant to Problem) |
|---|---|---|
| Cubing a number | Multiplying a number by itself three times ($\left.x^3 = x \times x \times x\right.$). | $\left(1.6\right)^3 = 1.6 \times 1.6 \times 1.6$ |
| Decimal Multiplication | Multiplying numbers with decimal points. The total number of decimal places in the product equals the sum of decimal places in the numbers being multiplied. | $1.89 \times 1.6$. $1.89$ has 2 decimal places, $1.6$ has 1. Product ($3.024$) has $2+1=3$ decimal places. |
| Algebraic Identity | An equation that is true for all possible values of its variables. | $\left(a+b\right)^3 = a^3 + b^3 + 3ab\left(a+b\right)$ |
Additional Information: Useful Algebraic Identities
Understanding algebraic identities can significantly simplify complex expressions and calculations. Here are a few other related identities:
- Cube of difference: $\left(a-b\right)^3 = a^3 - b^3 - 3ab\left(a-b\right)$
- Sum of cubes: $a^3 + b^3 = \left(a+b\right)\left(a^2 - ab + b^2\right)$
- Difference of cubes: $a^3 - b^3 = \left(a-b\right)\left(a^2 + ab + b^2\right)$
- If $a+b+c=0$, then $a^3 + b^3 + c^3 = 3abc$. This is a special case of the identity $a^3 + b^3 + c^3 - 3abc = \left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$. In our problem, if we let $a=0.9$, $b=0.7$, and $c=-1.6$, then $a+b+c = 0.9+0.7-1.6 = 1.6-1.6=0$. So $\left(0.9\right)^3 + \left(0.7\right)^3 + \left(-1.6\right)^3 = 3\left(0.9\right)\left(0.7\right)\left(-1.6\right)$. The original expression is $\left(1.6\right)^3 - \left(0.9\right)^3 - \left(0.7\right)^3 = -\left(\left(0.9\right)^3 + \left(0.7\right)^3 - \left(1.6\right)^3\right) = -\left(\left(0.9\right)^3 + \left(0.7\right)^3 + \left(-1.6\right)^3\right)$. Using the identity, this is $-\left(3\left(0.9\right)\left(0.7\right)\left(-1.6\right)\right) = -3\left(0.9\right)\left(0.7\right)\left(-1.6\right) = 3\left(0.9\right)\left(0.7\right)\left(1.6\right)$, which is the same result as Method 2. This confirms our approach.
These identities are fundamental tools for simplifying algebraic expressions and solving equations efficiently.
Important Questions from Decimals
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The value of \(\frac{48.3\times[(4.95)^2+4.95\times13.25]}{[(12.55)^2-(5.65)^2]\times19.8} \) is:
What is the value of x, if \(5\left( {1 - \frac{x}{5}} \right) - (5 - x) - \frac{1}{{200}}{\rm{of (20 - x) = 0}}{\rm{.08}}\) ?
The value of \(11.\overline{4}\) + \(22.5\overline{67}\) – \(33.5\overline{9}\) is:
