Find Sum of Fourth Powers Using Symmetric Identities

 

❓ Question

If:

$$a+b+c=1$$
$$ab+bc+ca=2$$

and

$$abc=3$$

then find:

$$a^4+b^4+c^4$$

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🖼 Question Image

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✍️ Short Concept

This question uses:

👉 Symmetric identities
👉 Sum of squares formula
👉 Algebraic expansion.

Main trick:

Convert higher powers into known symmetric forms.

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🔷 Step 1 — Find $a^2+b^2+c^2$ 💯

Using identity:

$$(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$$

Substitute values:

$$1^2 = a^2+b^2+c^2+2(2)$$
$$1=a^2+b^2+c^2+4$$
$$a^2+b^2+c^2=-3$$

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🔷 Step 2 — Square the Result

$$(a^2+b^2+c^2)^2 = (-3)^2$$
$$a^4+b^4+c^4 + 2(a^2b^2+b^2c^2+c^2a^2) =9$$

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🔷 Step 3 — Find $a^2b^2+b^2c^2+c^2a^2$

Square:

$$(ab+bc+ca)^2$$
$$= a^2b^2+b^2c^2+c^2a^2 + 2(ab\cdot bc+bc\cdot ca+ca\cdot ab)$$

Since:

$$ab\cdot bc+bc\cdot ca+ca\cdot ab = abc(a+b+c)$$

we get:

$$4 = a^2b^2+b^2c^2+c^2a^2 + 2abc(a+b+c)$$

Substitute:

$$abc=3,\quad a+b+c=1$$
$$4 = a^2b^2+b^2c^2+c^2a^2+6$$
$$a^2b^2+b^2c^2+c^2a^2=-2$$

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🔷 Step 4 — Final Calculation

Substitute into Step 2:

$$a^4+b^4+c^4+2(-2)=9$$
$$a^4+b^4+c^4-4=9$$
$$a^4+b^4+c^4=13$$

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✅ Final Answer

$$\boxed{13}$$

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⭐ Golden JEE Insight

For symmetric expressions:

$$a+b+c,\quad ab+bc+ca,\quad abc$$

always use identities first instead of expanding directly.

Most power-sum questions simplify very fast using standard formulas.