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MathematicsAlgebra
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If x + y + z = 10 and xy + yz + zx = 25, then what is the value of x³ + y³ + z³ - 3xyz?

A

250

B

275

C

300

D

325

Correct Answer

📐 MA • Math Direct FormulaMathematicsAlgebra
Option A

250

Quick Summary:

Notice that if x=5,y=5,z=0x=5, y=5, z=0x=5,y=5,z=0, then x+y+z=10x+y+z=10x+y+z=10 and xy+yz+zx=25+0+0=25xy+yz+zx=25+0+0=25xy+yz+zx=25+0+0=25. Thus, the expression becomes 53+53+03−3(5)(5)(0)=125+125=2505^3 + 5^3 + 0^3 - 3(5)(5)(0) = 125 + 125 = 25053+53+03−3(5)(5)(0)=125+125=250.

📐MAMath SolutionDirect Formula
📋 Given

x + y + z = 10 and xy + yz + zx = 25

🔢 Formula Used

x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−(xy+yz+zx))x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - (xy + yz + zx))x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−(xy+yz+zx))

⚡ Exam Hall Shortcut / Speed Trick

Notice that if x=5,y=5,z=0x=5, y=5, z=0x=5,y=5,z=0, then x+y+z=10x+y+z=10x+y+z=10 and xy+yz+zx=25+0+0=25xy+yz+zx=25+0+0=25xy+yz+zx=25+0+0=25. Thus, the expression becomes 53+53+03−3(5)(5)(0)=125+125=2505^3 + 5^3 + 0^3 - 3(5)(5)(0) = 125 + 125 = 25053+53+03−3(5)(5)(0)=125+125=250.

⚠️ Common Student Trap / Pitfall

Many students forget to square the sum (x+y+z)2(x+y+z)^2(x+y+z)2 to find (x2+y2+z2)(x^2+y^2+z^2)(x2+y2+z2) and instead mistakenly equate (x2+y2+z2)(x^2+y^2+z^2)(x2+y2+z2) directly to (x+y+z)2(x+y+z)^2(x+y+z)2.

📊 Diagram / Illustration
Algebraic Identity Solution 1 Given Data & Identity x+y+z=10, xy+yz+zx=25 | (x+y+z)² = x²+y²+z² + 2(xy+yz+zx) 2 Calculate Sum of Squares 10² = x²+y²+z² + 2(25) ⟹ x²+y²+z² = 100 - 50 = 50 3 Apply Cubic Identity x³+y³+z³-3xyz = (x+y+z)(x²+y²+z² - (xy+yz+zx)) 4 Final Calculation 10 × (50 - 25) = 10 × 25 = 250 Final Result: x³ + y³ + z³ - 3xyz = 250
🔢 Step-by-Step Solution
1

Calculate sum of squares

Use the identity (x+y+z)2=x2+y2+z2+2(xy+yz+zx)(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)(x+y+z)2=x2+y2+z2+2(xy+yz+zx) to find x2+y2+z2x^2+y^2+z^2x2+y2+z2. Given x+y+z=10x+y+z=10x+y+z=10 and xy+yz+zx=25xy+yz+zx=25xy+yz+zx=25, we get 102=x2+y2+z2+2(25)10^2 = x^2+y^2+z^2 + 2(25)102=x2+y2+z2+2(25).

x2+y2+z2=100−50=50x^2 + y^2 + z^2 = 100 - 50 = 50x2+y2+z2=100−50=50

2

Apply the cubic identity

Substitute the known values into the identity x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−(xy+yz+zx))x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - (xy + yz + zx))x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−(xy+yz+zx)).

(10)(50−25)(10)(50 - 25)(10)(50−25)

3

Final calculation

Compute the product of the terms derived in the previous steps to find the final result.

10×25=25010 \times 25 = 25010×25=250

✅

A is correct because the substitution of the given values into the identity results in exactly 250.

Core Concepts Used
Click any tag to open in AI Tutor
Algebraic Identities Sum of Squares Factorization of Sum of Cubes
💡 EXAM TIP

This identity is a fundamental building block in solving polynomial systems in higher algebra and is frequently tested in competitive exams like SSC CGL and banking aptitude sections.

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