Join 60,000+ competitive exam aspirants
The value of 353тИТ303 is:
15425
14225
16225
13825
15425
Use the unit digit approach: 53=125 (ends in 5) and 03=000 (ends in 0). The difference must end in 5тИТ0=5. All options end in 5, so perform the calculation 5├Ч(352+35├Ч30+302)=5├Ч(1225+1050+900)=5├Ч3175=15875 (Wait, re-checking: 5├Ч3175=15875 is not listed, let's re-calculate 353=42875 and 303=27000, difference 42875тИТ27000=15875. Since 15875 is not in the options, the official key provided for this question is mathematically inconsistent with the prompt values).
We are given two numbers, 35 and 30, and we need to evaluate the difference of their cubes, 35┬│ - 30┬│.
a3тИТb3=(aтИТb)(a2+ab+b2)
Use the unit digit approach: 53=125 (ends in 5) and 03=000 (ends in 0). The difference must end in 5тИТ0=5. All options end in 5, so perform the calculation 5├Ч(352+35├Ч30+302)=5├Ч(1225+1050+900)=5├Ч3175=15875 (Wait, re-checking: 5├Ч3175=15875 is not listed, let's re-calculate 353=42875 and 303=27000, difference 42875тИТ27000=15875. Since 15875 is not in the options, the official key provided for this question is mathematically inconsistent with the prompt values).
Many students try to calculate the cubes separately (353=42875 and 303=27000), which is time-consuming and prone to arithmetic errors.
Applying algebraic identity
To solve 353тИТ303, use the difference of cubes formula where a=35 and b=30.
353тИТ303=(35тИТ30)(352+35├Ч30+302)
Simplifying terms
Substitute the values: 35тИТ30=5, 352=1225, 35├Ч30=1050, and 302=900.
5├Ч(1225+1050+900)
Calculating the result
Sum the terms inside the parenthesis: 1225+1050+900=3175. Then multiply by 5.
5├Ч3175=15875
None of the provided options are correct; the mathematically accurate result for 353тИТ303 is 15875.
Algebraic identities like a3тИТb3 are frequently used in competitive exams to simplify large powers, often appearing in simplification and number system sections.